
Class A 
Book_ 



i2L- 



OopyrightN . 






COPYRIGHT DEPOSIT. 



HEAT ENGINEERING 



: hiiiiiiiiiiiiiiiiiiiiim Illliliill 

McGraw-Hill BookCompatiy 

Pu/iCis/ie/'s of3oo£§/br 

Electrical World TheEngineering andMining Journal 
Engineering Record Engineering News 

Railway Age Gazette American Machinist 

Signal Engineer American Engineer 

Electric Railway Journal Coal Age 

Metallurgical and Chemical Engineering P o we r 



HEAT ENGINEEKING 



A TEXT BOOK OF APPLIED THERMODYNAMICS 

FOR ENGINEERS AND STUDENTS 

IN TECHNICAL SCHOOLS 



BY 
ARTHUR M. GREENE, Jr. 

PROFESSOR OF MECHANICAL ENGINEERING, RUSSELL SAGE FOUNDATION 

RENSSELAER POLYTECHNIC INSTITUTE; SOMETIME JUNIOR 

DEAN, SCHOOL OF ENGINEERING, UNIVERSITY 

OF MISSOURI 



First Edition 



McGRAW-HILL BOOK COMPANY, Inc. 
239 WEST 39TH STREET, NEW YORK 

6 BOUVERIE STREET, LONDON, E. C. 
1915 



»« 



-Oft 



\ 



Copyright, 1915, by the 
McGraw-Hill Book Company, Inc. 



fS^ 



n 



THE MAPX.E PRESS YORK PA 



MAR 22 1,915 
©CI.A397220 



PREFACE 

For many years the author has given lectures supplementing 
the text-books used as a basis for a course in heat engineering. 
His aim in preparing this book has been to bring together his 
various notes with statements of the investigations and writings 
of others to make a complete treatment of the important phases 
of this subject. In doing this he has given credit to the authors 
and investigators quoted. Certain of the original sources have 
been quoted so that the student may learn the use of references. 
It is hoped that many studying this book will refer to these 
original papers. 

The work presupposes a course in theoretical thermodynamics 
such as that given in the treatises of Wood, Peabody or Goode- 
nough. Because' of the difference in symbols, nomenclature or 
point of view of various authors and to serve for reference or 
for the derivation of formulae used in the text, the first chapter of 
this book has been written. It is not intended that this chapter 
shall be used as a part of the course for it is an outline only of the 
thermodynamic theory. It should be used to give a review of the 
subject or as a basis for the f ormulae used. In shaping this chapter 
the author has been guided by his experience in teaching this 
subject from many texts. The treatment of availability and 
entropy has been based on the excellent work on thermodynamics 
by Goodenough. 

Numerical problems have been solved at various points in the 
text to illustrate the principles of the subject and to apply them to 
actual engineering work. The problems have been solved in 
detail to give the student one manner of attack as well as an order 
for the arrangement of computations for clearness. Unless the 
student can apply the various formulae and theories he has failed 
to attain that for which this book was written. In addition to the 
problems and solutions a series of questions on the various topics 
of the text and a set of problems illustrating their use have been 
placed at the end of each chapter. These may be used by the 
student in preparation of an assignment or by the teacher for 
blackboard recitations. 



vi PREFACE 

The author not only expresses his thanks to those whose works 
he has used and whose names he has placed in the first part of the 
index but to those whose writings he has studied as a student and 
teacher and whose work or whose view point he has absorbed. 
He especially thanks his wife, Mary E. Lewis Greene, for her aid in 
the preparation of manuscript, proof and final arrangement of 
work. A. M. G., Jr. 

Sunnyslope, Troy, N. Y., 
February 22, 1915. 



CONTENTS 

Page 

Preface ! v 

List of Symbols xi 

CHAPTER I 

Fundamental Thermodynamics 1 

Availability — Graphical Representation of Heat on P. V. Plane — 
Scale of Temperature — Carnot Cycle — Reversed Cycle — Maxi- 
mum Efficiency — Conditions for Availability of Heat — Second 
Law — Loss of Availability — Entropy — Entropy around a Cycle — 
Entropy Diagram — Characteristic Equations — Heat on a Path — 
Differential Equations and Relations — Perfect Gases — Differen- 
tial Heat Equations for Perfect Gases — Specific Heats — Isothermal 
and Isodynamic Lines — Adiabatic — Thermodynamic Lines — 
Equality of Gas Scale and Kelvin Scale — Heat Content of Gases — 
Entropy of Gases — Cycles and Cross Products — Saturated Steam 
and Other Vapors — Intrinsic Energy — Heat on a Path — Heat 
Content — Entropy — Differential Effect of Heat — Superheated 
Vapor — Specific Heats at Constant Pressure and k's — T-S and 
I-S Charts — Heat on Paths — Flow of Fluids — Throttling Action — 
Velocity of Various Substances — Discharge from Orifices. 

CHAPTER II 

Heat Engines and Efficiencies 62 

Method of Reporting Performance of Engines — High-speed Non- 
condensing Engine— High-speed Non-condensing Compound 
Engine — Locomobile Engine — Pumping Engine — Steam Turbine — 
Producer Gas Engine — Blast-furnace Gas Engine — Diesel Oil 
Engine — Results of Various Tests — Topics — Problems. 

CHAPTER III 

Heat Transmission .72 

Laws of Transmission — Mean Temperature Difference — Deter- 
mination of K-Steam Condensers — Ammonia Condensers — Brine 
Coils — Evaporators and Feed-water Heaters — Heat from Liquids 
to Liquids — Factor of Safety — Radiators — Heat through Walls 
and Partitions — Efficiency of Heat Transmissions — Topics — 
Problems. 

vii 



viii CONTENTS 



CHAPTER IV 

Page 

Air Compressors 118 

Various Types — Work of Compression — Effect of Clearance — 
Effect of Leakage — Volumetric Efficiency — Horse-power and 
Power of Motor — Temperature at the end of Compression — Com- 
pression Curve — Heat Removed by Jacket — Saving due to Jacket — 
Water Required for Jacket — Multistaging and Intermediate Pres- 
sures — Intermediate Temperatures — Heat Removed by Inter- 
cooler — Amount of Water for Intercooler — Water Removed in 
Intercooler — Effect of Leakage in Multistage Compressions — Dis- 
placement of Cylinders — Size of Inlet and Outlet Pipes and 
Valves — Percentage Saving due to Multistaging over a Single 
Stage — Unavoidable Loss of Compression on Two Stages — 
Work on Air Engine — Loss due to Cooling after Compression — 
Unavoidable Loss due to Expansion Line — Loss of Pressure in 
Pipe Line — Loss due to Leakage from Transmission Line — Loss 
due to Throttling — Gain from Preheating— Power and Displace- 
ment of Air Engine or Motor — Fan Blowers — Governing — Motors 
— Losses in Transmission — Logarithmic Diagrams — Table of 
Power — Topics — Problems . 



CHAPTER V 

The Steam Engine 167 

Action of Steam Engine — Cycle of Steam Engine — Heat and Effi- 
ciency of Cycle — Steam Consumption — Temperature-Entropy 
Diagrams — Effects of Changes — Calorimeters — Analysis of Tests — 
Hirn's Analysis — Temperature Entropy Analysis — Missing Quan- 
tity and Initial Condensation — Experiments of Effect of Cylinder 
Walls — -Valve Leakage — Steam Consumption by Clayton's Method 
— Values of N for Expansion Lines — Expansion Lines — Use of Rec- 
tangular Hyperbola — Construction of Expansion Curves — Mean 
Effective Pressure — Real and Apparent Ratio of Expansion — 
Stumpf Engine — Size of Engines — Best Point of Compression — 
Speed — Locomobile — Topics — Problems. 



CHAPTER VI 

Multiple Expansion Engines 240 

Action — Combined Cards — Computation of Cards for Construction 
— Equivalent Work done by One Cylinder — Determination of 
Relative Sizes of Cylinders — Jacketing — Reheaters — Governing 
— Bleeding Engines or Turbines — Regenerative Engines — Testing 
and Analysis — Binary Engines — Topics — Problems. 



CONTENTS IX 

CHAPTER VII 

Page 

Steam Nozzles, Injectors, Steam Turbines 265 

Nozzles — Injectors — Steam Nozzle Velocity — Water Velocities — 
Theory of the Injector — Heat Equation of the Injector — Steam 
Weight — Steam Nozzle — Combining Tube — Delivery Tube — 
Density — Impact Coefficient — Injector Details — Steam Turbines 
— Action of Jets — Maximum Efficiency — Types of Turbines — 
Efficiency — Similarity of Action of Turbine and Engine — Compu- 
tations and Design — Reheat Factor — Combined Engine and Tur- 
bine — Allowance for Change of Running Conditions — Combined 
Engine and Turbine — Topics — Problems. 

CHAPTER VIII 

Condensers, Cooling Towers and Evaporators 333 

Types of Condensers — Pressures in a Condenser — Condenser 
Design — Problems — Cooling Towers — Size of Tower and Mats — 
Spray Nozzles and Ponds — Accumulators and Evaporators — 
Double Bottoms — Temperature of Boiling — Topics — Problems. 

CHAPTER IX 

Internal Combustion Engines and Combustion 359 

Development of Engine — Otto Cycle — Air Standard Efficiency — 
Atkinson Cycle and Diesel Cycle — Actual Engines — Governing — 
Ignition — Heat Transfer to Walls — Fuels — Combustion of Fuels — 
Gas Composition — Problem — Temperatures at Corners of Card — 
Adiabatics — Efficiency Change — Temperature after Explosion — 
Temperature Entropy Diagram — Logarithmic Diagram — Test of 
Gas Engine — Combustion of Fuels for Boilers — Surface Combus- 
tion — Topics — Problems. 

CHAPTER X 

Refrigeration 411 

Air Machines — Problem — Effect of Clearance and Friction Mois- 
ture — Vapor Refrigerating Machines — Effect of Substances — Ab- 
sorption Apparatus — Properties of Aqua Ammonia — Investigation 
of Absorption Apparatus — Topics — Problems. 

Index 455 



SYMBOLS USED IN TEXT 

A = -j = Tyf^ = const, to change foot pounds to B.t.u. 

A = constant in Reynold's equation. 

a = cleaness factor Orrok's formula. . . 

a = coefficient of heat transmission at surface. 

a = coefficient in various equations or length of line. 

pV 

B = gas constant = ^p = foot-pounds per deg. per pound. 

B = constant in Reynold's equation. 

b = constant in various equations or lengths of lines. 

b = fraction of volume at compression in Mark's formula. 

b.h.p. = brake horse-power. 

B.t.u. = British thermal units. 

C = constant in equation. 

c = coefficient of conduction or heat transmission. 

c = specific heat of water or length of line. 

c' = specific heat of liquid. 

c" = specific heat of saturated steam 

C P = specific heat at constant pressure of 1 cu. ft. 

c p = specific heat at constant pressure. 

C v = specific heat of constant volume of 1 cu. ft. 

c v — specific heat of constant volume. 

D = displacement of cylinder in cubic feet or length of card in inches. 

d = diameter of cylinder or pipe in feet ( or inches). 

d' = hydraulic radius. 

d = constant in equation for air movement in heat transmission or 

length of line. 

D.H.P. = delivered horse-power. 

e = area of elementary strip in Kelvin diagram. 

e = fraction of steam at cut-off. 

e ■= weight of evaporation in evaporator. 

e = constant for material in formula for heat transmission. 

F = area or surface in square feet (or square inches) 

F = Maxwell's thermodynamic potential. 

/ = leakage factor, diagram factor, or percentage friction. 

/ = a/1 — y = friction factor for velocity. 

G = lbs. of water per minute or per pound of steam. 

g = acceleration of gravity. 

H = heat from internal friction in B.t.u. 

H = heat in 1 cu. ft. of gas. 

h — feet head. 

h' = lift of injector in feet. 

H.P. = horse-power. 

/ = heat content of M lbs. of substance. 



xii SYMBOLS USED IN TEXT 

i = heat content of 1 lb. of substance. 

i.h.p. = indicated horse-power. 

J = Joule's equivalent = 778 = constant to change B.t.u. to foot- 
pounds. 

K = coefficient of conduction and coefficient of various equations. 

k = ratio of c p to c v or coefficient of equation. 

K2 = clearance factor. 

Kw = kilowatts. 

L = stroke in feet. 

I = percentage clearance or periphery of valves. 

I = length of path or thickness of material in feet. 

lp and l v = latent heats. 

M = mass of substance in pounds. 

m = mass of 1 cu. ft. of substance in pounds or constant in equations. 

M a = apparent weight of steam per card, indicated. 

Mi = apparent weight of steam per card, indicated. 

m f = weight of steam per cubic foot of displacement. 

M m = missing weight of steam. 

Mo = clearance steam per card. 

m = weight of cubic feet of steam. 

m.q. = missing quantity in per cent, of steam used. 

M mai = steam used by engine. 

N = revolutions per minute. 

n = exponent of volume terms in equations. 

n = no. of molecules. 

n = heat of expansion. 

o = heat of pressure change. 

P = total force or pressure. 

P = mean effective pressure on piston. 

p — normal pressure in pounds per square foot (pounds per square 
inch or millimeter of mercury). 

p m = friction factor in pipes. 

Q = heat added to M lbs. in B.t.u. 

q = heat per pound of substance or heat of solution of aqua ammonia. 

q' = heat of liquid, at outlet, q' ; at inlet q\. 

q" = total heat in 1 lb. of saturated steam. 

q"> = total heat of 1 lb. of superheated steam. 

R = universal gas constant = 1544 ft.-lbs. per deg. per molecule. 

R = total ratio of expansion. 

r = heat of vaporization. 

r = ratio of expansion. 

— = cut-off. 



7p = entropy of vaporization. 

r r — real ratio of expansion. 

r.e. = residual energy. 

R.H. — reheat factor. 

s = stroke in feet. 

s = normal cylinder surface per cubic foot of displacement. 



SYMBOLS USED IN TEXT xiii 

s' = Entropy of 1 lb. liquid. 

s" = total entropy of 1 lb. steam. 

S 2 — Si = change of entropy for M lbs. of substance. 

«2 — «i = change of entropy per pound of substance. 

T = temperature in degrees from absolute zero. 

T = temperature difference in Heck's formula. 

t = temperature from F. or C. zero. 

Tsat = temperature of boiling. 

Tsoi = temperature of boiling of aqua ammonia. 

U = intrinsic energy of M lbs. of substance in foot-pounds. 

u = intrinsic energy of 1 lb. of substance. 

V = volume of M lbs. of substance in cubic feet. 

v = volume of 1 lb. of substance in cubic feet. 

v' = volume of 1 lb. of liquid. 

v" = volume of 1 lb. of dry saturated steam. 

W = work in foot-pounds for M lbs. of substance. 

w = work for 1 lb. of substance. 

w = velocity of substance in feet per second. 

w a = actual velocity of jet in feet per second. 

ivb = velocity of blade in feet per second. 

w r = relative velocity of jet. 

wt. = molecular weight. 

x = quality or dryness factor = pounds of steam per pound of 

mixture. 

x = per cent, of NH 3 in 1 lb. of solution. 

x = point of compression, distance moved and variable length of 

injector. 

x" = distance on card occupied by boiler steam. 

y = friction factor for velocity. 

y = amount of liquor to produce given change in concentration. 

z = pounds of water per pound of steam in injector. 

a = acceleration. 

a = temperature coefficient. 

a = angle of actual velocity. 

/3 = angle of relative velocity. 

7 = angle on I-S diagram. 

5 = angle on I-S diagram. 

A = relative density in injector. 

At = temperature difference, 

e = base of natural logarithms. 

77 = efficiency or over-all efficiency. 

771 = Carnot efficiency. 

772 = type efficiency. 

773 = theoretical efficiency. 
7?4 = practical efficiency. 

775 = actual efficiency. 

776 = mechanical efficiency. 
7? e = electrical efficiency. 
Tjfc = kinetic efficiency. 



xiv SYMBOLS USED IN TEXT 



T\ m = mechanical efficiency. 

r) n = nozzle efficiency. 

7] r = refrigerative efficiency. 

77s = efficiency of stage. 

■qt = over-all efficiency. 

t) w = efficiency of weight. 

6 = temperature of material. 

X = coefficient of conduction of gas. 

/j, = materials factor in Orrok's formula. 

p = internal heat of vaporization. 

p = relative humidity. 

p = steam richness factor in Orrok's formula. 

<t> = Maxwell's thermodynamic potential. 

<f) = coefficient in Nicolson's formula of heat transmission. 

\p — external work in making steam. 



HEAT ENGINEERING 

CHAPTER I 
FUNDAMENTAL THERMODYNAMICS 

Heat is a form of energy and as such it may be measured in 
any unit of energy. The customary unit is the British thermal 
unit, B.t.u., although heat may be measured in foot-pounds, ergs, 
joules, calories, horse-power hours, kilowatt hours or any other 
unit of energy. The B.t.u. is 3^80 of the amount of heat neces- 
sary to raise the temperature of 1 lb. of water from 32° F. to 
212° F. 

Heat energy is a form of energy due to a vibration or motion 
of the molecules of a body. It may be produced by the transfor- 
mation of other forms of energy into heat, as when mechanical 
energy of a moving train is changed into heat by brakes. 
When other forms of energy are produced from heat energy, how- 
ever, it is found that only a portion of the heat energy may be 
transformed. This leads at once to the separation of energy 
into two classes: high-grade energy and low-grade energy. 
High-grade energy is any kind of energy which may be completely 
changed into any other form. Mechanical and electrical energy 
are examples of this while heat energy is low grade since it cannot 
be changed completely into any other form of energy. For this 
reason all of the heat energy in a body is not available for trans^ 
formation. The fractional part of the heat energy which is 
available for transformation into another form is known as the 
availability of heat energy. If Q heat units are in a system 
and if part of these are changed into another form of energy and 
Q' heat units remain in the system, then Q — Q' heat units must 
have been changed. If this is all that could have been changed, 

Q-Q' 

— ■« — represents the fractional part or the availability. It has 

been shown by Joule and others that when mechanical work is 
changed into heat or when heat is partially changed into mechan- 
ical energy there is a definite relation between the work and the 

1 



2 HEAT ENGINEERING 

heat in the first case and between the disappearance of heat and 
the work produced thereby in the second case. This statement 
is known as the first law of thermodynamics. The numerical 
relation between heat and work is known as Joule's equivalent, 
/. It is equal to 778 ft .-lbs. 

1 B.t.u. = 778 ft.-lbs. (1) 

JQ = W (2) 

For convenience the reciprocal of J is used at times, the symbol 
for this being A. 

A = A (3) 

Q = AW (4) 

Q = Heat in B.t.u. W = Work in ft.-lbs. 

In French units the value of J is 426 kg. m. = 1 kg. deg. calorie. 
Now when heat is added to a body it may increase the internal 
energy of that body and do external work or 

JdQ = dU + dW (5) 

This formula is a mathematical statement of the first law of 
thermodynamics which is only one aspect of the law of the con- 
servation of energy ; heat added to a body increases the energy 
of that body and does external work. U is the symbol used to 
indicate internal energy or intrinsic energy of a body. The capi- 
tal letters Q, U, W, refer to a body of any weight. If a body of 
unit weight is considered, small letters are used. 

Jdq = du + dw (6) 

Now in the above equation any one of the differential quanti- 
ties may be zero or have any sign. Thus if dq = 0, no heat is 
added and such action is known as adiabatic action. If du = 0, 
there is no change in intrinsic energy and the action is known as 
isodynamic action, while if dw = 0, no external work is done, and, 
as will be seen later, there is no change in volume. The signs 
of these terms may be anything and the equation is true. This 
equation is one of the important ones in thermodynamics. If at 
any time the change of intrinsic energy is known on any path 
or during any change of state and if the work is known during 
that change, then the sum of these two will be the heat required 
during this change. If positive, heat must be added; if negative, 
heat must be abstracted. 



FUNDAMENTAL THERMODYNAMICS 3 

Now p represents the pressure on unit area (1 sq. ft.) of the 
surface of the body considered and it is assumed uniform and nor- 
mal to the surface surrounding the body. If then the total area 
of the surface of the body isF,pF is the total normal force on the 
body. If this surface is moved through the normal distance dx, 
the work done is 

pFdx = dW 



but since 



or 



Fdx = dV 

pdV = dW (7) 



pdv = dw 
The equation (6) may therefore be written 

Jdq = du + pdv (8) 

In this equation du must be measured in the same units as pdv, 

namely foot-pounds, and dq on account of the constant J must be 

measured in heat units. The total heat is the integral of dq or 

r*b r** r*b 

J I dq = I du + I pdv 

«y a U a U a 

The internal energy of a body depends on the condition of the 
body and not on the method of bringing it to that state and conse- 
quently the change in intrinsic energy, which is I du, does not 

depend on the manner in which one state is changed to the other 
but merely on the states at the beginning and the end of the 
operation. Hence, if the intrinsic energy at state a is u a and at 
state b is Ub, the value of the integral du between these two 
states is u b — u a . This is true whatever the path may be. 

'b 

pdv represents the area beneath a curve on the pv plane and 



1 



this area depends on the path considered. Hence the value of 
this integral can only be told after the curve or path is known. 
The integral of dq depends therefore on the path since, although 

J du is independent of the path, J dw or I pdv does depend 

on the path. A differential whose integral does not depend on 
the path is called an exact differential because it can be inte- 



4 HEAT ENGINEERING 

grated directly. Hence there must be some functional relation 
between the independent variables in regard to which it is being 
integrated, otherwise a path would have to be fixed or some rela- 
tion would have to be known to give one variable in terms of the 
other to make the integration possible. Thus 



I ydx = I dX 



is not an exact differential since there must be some known rela- 
tion between y and x before the integration can be performed. 



Now I dx = x and I {xdy + ydx) = 



xy 



are each directly integrable and their definite integral values 
depend only on the limits and not on the path between the limits 
and for this reason each quantity behind the integral sign is an 
exact differential. It. is seen that if exact differentials be inte- 
grated around a closed path their value would be zero since both 
limits are the same. This is one way of telling whether or not a 
differential is exact. If the integral on every closed path is 
zero, the differential is exact. There is one other criterion by 
which an exact differential may be told or a relation which must 
exist if a differential is exact, and if known to be a function of 
two independent variables. Thus if 

dX = Mdx + Ndy 

dX is an exact differential if 

by ~ bx W 

If however dX is known to be exact by some other method, 
then 

8M , ^ ^N w\ 

-r— must equal -r— (9 ) 

by ^ bx 

The reason for this is seen from the fact that if dX is exact there 
must be some functional relation 

X = f(xy) 
hence 

d)[ = -t— dx + -7— dy 

bx by 



FUNDAMENTAL THERMODYNAMICS 



now 

but 

hence 



This relation between the differential coefficients M and N 
may be used to determine whether or not the differential is exact, 
or if known to be exact, the equality of the partial derivative of 
M and N may be used to determine new relations as will be seen 
later. 

The fundamental equation may be written 



bX _ 
8x 


M and — 
by 


= N 


8 2 X 


8 2 X 




dx8y 


bybx 




b 2 X 


bM b 2 X 


bN 


bxby 


by bybx 


bx 



Jq = u b — u a + 



P 

I pdv 



If q = 0, the line is an adiabatic and 



U a — U b 



P 

I pdv 

*J a 



(10) 



(11) 



This means that work during an adiabatic change is equal 
to the change of intrinsic energy, or work is done at the expense 
of intrinsic energy. If ab oo , 
Fig. 1, represents an adiabatic 
path of a substance on the pv 
plane, the area lab2 represents 
the change of intrinsic energy 
from a to 6. If this is car- 
ried out to infinity there can 
be no further area under the 
curve and the area from a to 
this point must represent the 
total intrinsic energy at a since 
no heat has been added and 

work has been done to the point of zero temperature. Although 
this area is infinite in extent it is not infinite in value since the 
amount of energy in a body cannot be infinite. This is seen to 
be true mathematically since the curve approaches the v axis rap- 




v 2 
Fig. 1. — Adiabatic on the pV plane. 



6 



HEAT ENGINEERING 



idly and the height of the curve is practically zero after a short 
distance to the right. 

If u does not change on the curve (the isodynamic) the heat 
added is equal to the work done. 

Jq = I pdv (12) 

I pdv = 0, or dv = 0, and v = constant. 

Jq = u b — u a (13) 



If there is no work. 
In this case 



GRAPHICAL REPRESENTATION OF HEAT ON P.V. PLANE 

In Fig. 2 let ab be any path. Draw from a and b two adia- 
batics to infinity and from a draw the isodynamic until it strikes 
the adiabatic from b at c. 




12 v 

Fig. 2. — Graphical representation of heat added. 



Now la oo = u a 






2b co = u b 






lab2 = 1 pdv 






26 oo + l a &2 = 


la&co 


f 

= u b + 1 pdv 


lab oo — la oo = 


oo ab oo 


r 

= Ub + 1 pdv - 

t/ a 

u b — u a -\- 



p 

t/ a 



J? 



FUNDAMENTAL THERMODYNAMICS 7 

This gives the important relation that the area on the pv 
plane between any path and the two adiabatics from the ex- 
tremities of the path to infinity is equal to the heat added on the 
path. 

This area is infinite in extent and consequently cannot be 
represented graphically. To make it finite and definite, the iso- 
dynamic ac was drawn. Now u a = u c . Hence 3c °° = la °° and 

lab co — 3c oo = Jq = labcS 

or the area beneath any path added to that beneath an adiabatic 
from the second point of that path to the intersection of the 
adiabatic and the isodynamic from the first point is equal to the 
heat added on the path. It will be seen that 



lab2 
and 



f 

= I pdv 



26c3 = U b — U c = Ub — u a 

This latter statement should be clear since u c = u a . 

SCALE OF TEMPERATURE 

When heat is added to a body this body comes into a state 
in which it will transmit heat to another body with which it had 
been previously in contact. This ability to transmit heat to 
another body is determined by a property which is termed tem- 
perature. The temperature is measured by allowing an instru- 
ment to come into thermal equilibrium with the body and then 
noting the effect on the instrument. If one body is at a higher 
temperature than another it will transmit heat to that body. 

Suppose ab becomes a line of constant temperature T, such a 
line is called an isothermal (Fig. 3). If a°° and boo are adia- 
batics, the area ooaboo is the heat added on this line ab and is 
finite in value since it is equal to lab2 + 2b oo — la oo and each 
of these is finite. If q ab is divided by T the result 

m may be called e. 

If the isothermal a'b' is drawn below ab so that the area abb' a' 
is equal to e and then a second isothermal a^b" is so placed that 
a'b'b"a" is equal to e and this is continued, it is found that there 



8 



HEAT ENGINEERING 



will be T such isothermals drawn. Each isothermal is so drawn 
that the small areas are equal and these isothermals determine 
definite temperatures. These temperatures form a scale and 
since this results from considerations of absolute units of work 
and does not depend on any particular substance it is called 
the Kelvin absolute scale of temperature. 

The ordinary scale of temperature was first determined by the 
effects of heat on mercury or other fluids but on account of changes 
in the rate of expansion of these fluids by which the temperatures 
were measured, gases were suggested as the media for measure- 
ment and hydrogen was found to be the most constant for all 
measurements. There are two scales in common use in this 
country, the Fahrenheit in which the temperature of melting ice 
is 32° and that of boiling water under standard conditions is 




x v 2 
Fig. 3. — Isothermals for Kelvin's Scale. 



212°, and the Centigrade scale on which the temperature of 
melting ice under standard conditions is 0° and that of boiling 
water under standard conditions is 100°. Neither of these start 
at the true zero of temperature. In Centigrade units the true 
or absolute zero is at — 273° C. and in Fahrenheit units the zero 
is at — 459.6° F. The relative size of degrees in the two systems 
is given by 

1° C = %° F 

It will be shown that Kelvin's Absolute Scale is the same as 
that determined by the hydrogen thermometer. 

In looking at the diagram from which the Kelvin scale is 
fixed it will be seen that the area beneath any of the isothermals 
and the adiabatics is equal to the heat on the line and that 
these areas are equal to T times the unit area e. Hence the 



FUNDAMENTAL THERMODYNAMICS 



9 



heat added on any isothermal between points of intersection 
with two adiabatics is proportional to the temperature on that 
isothermal. This is an important consideration. 



Q =eT 



(14) 



is the expression for the heat added on an isothermal of tem- 
perature T between two adiabatics and 

Q' = eT' 

is true at temperature T' between the same two adiabatics. 
This leads to the efficiency of the Carnot cycle. 

CARNOT CYCLE 

Carnot proposed a cycle made up of isothermal and adiabatic 
lines in a peculiar form of engine. Imagine a perfect non-con- 
ducting cylinder with a perfectly conducting head containing a 
non-conducting piston together with two bodies S and R of infinite 























fe 


c 






$1 p- 










T 1 


E 


mm 


p 


///////// 


m 


R 

T 2 



Fig. 4. — Carnot engine. 



capacity for heat and a non-conducting plate P (Fig. 4). The 
fact that S and R are of infinite capacity means that if a finite 
amount of heat is added or taken away from them their tem- 
peratures will not change. The body S being at a higher tem- 
perature, T\, than that of the body R at T2, is known as the 
source while R is called the refrigerator. 

If now the cylinder C is placed on the body S and the pressure 
on the piston rod is made a differential amount less than that 
exerted by the substance within the cylinder on the piston, the 
piston will be driven upward. This would mean that work is 
done by the substance within the cylinder and this would im- 
mediately cause its temperature to fall. On account of the 



10 HEAT ENGINEERING 

perfect conductor forming the head of the cylinder, heat will 
flow at once into the substance and keep its temperature con- 
stant, so that if this operation is allowed to go on, a certain 
amount of external work would be done, a certain amount of 
heat would be abstracted from S and the substance in the 
cylinder would be left in a given condition. If the pressure on 
the piston rod in this second condition is kept at a differential 
amount above the pressure exerted on the piston by the sub- 
stance, the piston will move downward compressing the sub- 
stance within and thus doing work upon it, which tends to in- 
crease the temperature causing a flow of heat into S the tem- 
perature of which cannot change. After the starting point 
is reached the substance within the cylinder is in its original 
condition with a restitution of the heat taken from S and the 
development of the first external work on the substance within. 
These two actions out and back are isothermal and because (a) 
they can be imagined to take place in either direction, with 
(6) external conditions differing by an infinitesimal quantity 
and because (c) all things external and internal can be restored 
to the initial conditions by coming back over the path to the 
original condition by a reversal of directions; such action is 
known as reversible action. These three conditions must hold 
for all reversible actions. 

If instead of returning to the original condition after the addi- 
tion of Qi heat units from S, the cylinder is slipped to the non- 
conducting plate and then the substance within the cylinder is 
allowed to expand by causing the pressure on the rod to be a 
differential amount less than the pressure on the piston, the 
substance within can receive no heat and the expansion will be 
adiabatic. The external work will be done at the expense of or 
by the intrinsic energy in the substance. Of course the tem- 
perature within the body must fall and when this reaches the 
temperature of R, (T 2 ), the operation is stopped and the cylinder 
is put into communication with R. 

If before the cylinder is put into communication, the force 
on the piston rod is increased an infinitesimal amount, the sub- 
stance within the cylinder will be compressed by external energy 
applied and this operation will be adiabatic, taking place in the 
reverse direction over the same path. It is seen that this ful- 
fills the three conditions mentioned above and hence the adiabatic 
line is reversible. 




V 
Fig. 5. — Carnot cycle. 



FUNDAMENTAL THERMODYNAMICS 11 

When the cylinder is connected with R and the force on the rod 
is increased by a differential amount, the substance within is 
compressed. This tends to increase the temperature which 
causes a flow through the perfect conductor into the body of in- 
finite capacity. This causes reversible isothermal compression. 
Suppose that this is continued until Q 2 heat units are abstracted 
to such a point that when the cylinder is removed to P and 
adiabatic compression to the temperature T\ occurs, it will be 
in its original condition. This is shown on the pv plane by Fig. 
5. In this 1-2 is an isother- 
mal of temperature Ti on 
which Qi heat units have been 
added. Point 2 is arbitrary. 
2-3 is an adiabatic on which 
no heat has been added. The 
point 3 is fixed by its tem- 
perature T 2 at which the adia- 
batic expansion must stop. 
3-4 is an isothermal of tem- 
perature T 2 on which Q 2 heat 
units have been abstracted and on which the point 4 is fixed so 
that the adiabatic 4-1 on which no heat is added will end at 
the original point 1. This is known as the Carnot cycle. 

REVERSED CYCLE 

There is no reason why these operations could not occur 
in the reverse direction. In this case Q 2 heat units would be 
taken from R and Qi heat units would be given up to S. 

Consider the path 1, 2, 3, 4, 1 on which the total heat added is 

Qi + - Q 2 + = Qi - Q 2 
but 

Q = U b - U a + AW 

and on the closed path 

U b = U a 

• 

since the operation is brought back to the original point. 
Hence 

Q 1 -Q 2 = AW 

If this takes place in the opposite direction 
Q 2 - Q 1 = - AW 



12 HEAT ENGINEERING 

or work AW must be done from the outside so that Qi heat 
units may be discharged into the source when a smaller quantity 
Q 2 has been added from the refrigerator. 

Of course these must be so since if the substance within the 
cylinder has been brought back to its original condition and Qi 
heat units have been added while Q 2 have been abstracted, the 
difference Qi — Q 2 must have gone into external work. In 
any case since 

J JdQ = U 2 - Vi + jdW 

j jdQ = CdW (15) 

if U 2 — Ui = 0. In any closed cycle (a path ending at the 
point from which it starts) the algebraic sum of the heats on all 
paths of the cycle is equal to the work done. This is regardless 
of the reversibility of the path. 

The heat supplied on any cycle is Qi and hence its efficiency is 

AW = Q^Q, = Eff (16) 

For the Carnot cycle this may be simplified, since the heat 
transfers are on isothermals limited by the same two adiabatics. 
In this case 

Q 1 = eT 1 [See Eq.(14)] 

Q2 = eT 2 
• t?& e(Ti — T 2 ) T\ — T 2 , ,« 

• • Eff - = w l = ~^~ (17) 

This is the value of the efficiency of the Carnot cycle and it is 
the expression for availability as will be seen later. Hence be- 

Ti — T 2 

tween the limits 7\ and T 2 , heat has an availability — ^ 

1 1 

if used on a Carnot cycle. 

MAXIMUM EFFICIENCY 

Suppose there is a cycle of efficiency greater than that of 
the Carnot cycle. Call this cycle, cycle r. If there is a source 
and a refrigerator of temperatures T\ and T 2 , the Carnot cycle 
and other cycle may be operated between these, and the efficiencies 
will give the inequality 

Qlc — Qlc Qlr — Qir 

Qu < Qlr 



FUNDAMENTAL THERMODYNAMICS 13 

Suppose these two engines be connected to the same shaft 
and the efficient engine be used as the driver to drive the 
Carnot engine in a reversed direction. In this case Q\ r will be 
taken from S and Q lc will be given to the source. Neglecting 
friction the work required by one will just equal the work 
developed by the other. 

Qlc — Q^c = Qlr ~ Qlr 

Since the inequality is true 

Qlc > Qlr 

or more heat is being added to the source than is taken from it. 
Since the two engines coupled to the source and refrigerator re- 
ceive no energy or give no energy to the outside (one drives the 
other) the heat going to the source must come from the re- 
frigerator, and there exists something with no connection with 
outside systems which causes heat to flow from a point of low 
temperature to a point of higher temperature when placed be- 
tween the two points. This is unthinkable from all experience, 
hence Q u cannot be greater than Q lr and the efficiency of the 
cycle cannot be greater than that of the Carnot cycle. Hence 
the Carnot cycle is as efficient as any cycle. 

Suppose now the cycle is reversible, in which case the Carnot 
cycle might be used as the driver if of greater efficiency than that 
of the other cycle. 

Qlc — Qzc Qlr — Qlr 

Qlc Qlr 

Qlc — Qlc — Qlr ~ Qlr 
Qlr > Qlc 

Or the amount received by the source from the reversed cycle 
is greater than that taken by the direct Carnot engine. This 
cannot be so and hence a Carnot cycle cannot have a greater 
efficiency than that of the reversible cycle. But none can be 
greater than the Carnot. 

Since the efficiency of the cycle r cannot be greater and can- 
not be less than that of the Carnot cycle it must be equal to it 
and hence all reversible cycles have the same efficiency, which is 
equal to that of the Carnot cycle. All that can be said of non- 
reversible cycles is that they cannot have greater efficiencies than 
that of the Carnot cycle. 



14 HEAT ENGINEERING 

This proof can be used to show that the efficiency is inde- 
pendent of the medium used, for if one substance gave a higher 
efficiency it could be used on a Carnot cycle as a driver while the 
other substance would be used as the medium of a reversed 
Carnot cycle. The reasoning would lead to the same absurd 
result and hence all substances will give the same efficiency theo- 
retically when operating between the same source and refrigerator. 

The Carnot efficiency is the maximum efficiency aDd if the 
limiting temperatures are T\ and T 2 

is the maximum availability. Being the maximum it is that 
which should be obtained in practice theoretically and hence 
it is called simply the availability. 

CONDITIONS FOR AVAILABILITY OF HEAT 

This leads to the observation that the only way in which heat 
can be available is to have a difference in temperature necessi- 
tating two bodies at different temperatures and a substance 
which can receive and discharge the heat. The three systems 
are necessary to make heat available. 

(rp rp \ 
— Sp jQi are available for useful work 

T 
and Qitft must be rejected. Of course this rejected heat or 

1 1 
wasted heat so far as the work AW is concerned may be used 
for some valuable purpose, as when it is rejected from a steam 
engine and used for warming buildings or for heating water for 
a wash house. 

This leads to the Second Law of Thermodynamics: 

SECOND LAW 

"It is impossible by means of a self-acting machine unaided by any 
external agency to convey heat from one body to another at higher 
temperature" or "no change in a system of bodies that can take place 
of itself can increase the available energy of the system." When the 
matter is discussed in the manner given above, this law results in the 
expression 

rp rp 

— ~ — = Availability or Carnot Efficiency, 



FUNDAMENTAL THERMODYNAMICS 15 

is the conclusion from this law when the gain in unavailable energy is 
considered, as will be seen later. Each of the above expressions 
results from the second law of thermodynamics. 

Having seen two paths which are reversible, the isothermal 
and the adiabatic, it may be well to consider some changes which 
are irreversible and for that reason cannot truly be represented 
on the pv plane. 

LOSS OF AVAILABILITY 

Suppose heat flows by conduction from one body of high tem- 
perature to one of low temperature. Such action cannot possibly 
be assumed to take place in the reversed direction and therefore 
on account of the first requirement it is non-reversible. 

If a gas is assumed to discharge freely from a vessel of pressure 
pi to a place of distinctly lower pressure p 2 , such action could not 
be assumed to take place in an opposite manner, and lastly if 
work be changed into heat as by friction this could not be as- 
sumed to take place in a reverse direction. 

In all of these there is a loss of availability. 

In the first the loss of available heat would be the difference 
of the available heats at the two temperatures. This is 

if To is the lowest possible temperature, and T\ and Ti are the 
temperatures of the two sides of the conducting surface. 

In the second case the pressure has been brought nearer to 
the pressure of the surrounding medium and hence by the time 
it is brought to this pressure by adiabatic expansion the tem- 
perature will not be so low as it could have been before. Hence 



4 - S] > 4 - w] 



Since 

To < To' 
Ti is almost equal to T/ 

Loss in available heat = T ^ To' ytr 

i 1 1 1 

In the third case if A IF units of work are changed into heat 

of temperature T\ the available part of this is 



ATf(l-J)or (l-J- : ) 



16 HEAT ENGINEERING 

All of it was available as mechanical energy before the change, 
hence 

represents the loss of available energy. 

In the reversible changes or on the reversible cycles the un- 
available energy before the operation was 

Qi Yl 

and this is the amount rejected by the reversible cycle. Hence 
in this case there is no increase of unavailable energy. 

ENTROPY 

Goodenough points out that in all of these cases the non- 
reversible changes have brought about a loss of available energy 
and that these expressions are all of the form 



'°(rr) 



He then says that the quantity which when multiplied by To, 
the lowest available temperature, gives the increase of un- 
available energy due to a non-reversible or other change is called 
the increase of entropy. 

This quantity is of the form -^ and refers to one system or to 

several systems. If one system alone is considered there may 
be an increase or decrease of entropy due to the change of 
heat, while if two systems are considered reversible changes lead 
to no change in unavailable energy and hence no change of 
entropy, while non-reversible changes lead to an increase of un- 
available energy and consequently an increase of entropy. 
Thus if the source alone is considered when Qi heat units are 
given up to the cylinder of the Carnot engine, there is an un-> 

available amount of heat T ^r taken from this source and 

i i 

given to the medium in the cylinder. There is a decrease of 

entropy of y^- for the source but that of the medium is increased 
1 1 

by -jfi- and the sum of these two is zero. This is true if the cycle 
I i 



FUNDAMENTAL THERMODYNAMICS 17 

is worked in the reverse direction. If there is conduction of 

heat the loss of entropy will be —. while the gain of entropy will 

1 1 

be^-. This latter is greater since T 2 is less than T\. Hence 

1 2 



is a gain in entropy 








Qi 


Qi 




T 2 


Ti 



when the irreversible change takes place. This is the only 
direction in which this operation can take place, while in re- 
versible action the operation may take place in either direction. 
Hence if all bodies or systems taking place in a change be con- 
sidered together, a process which will take place of itself will be 
accompanied by an increase of entropy. If there is no change 
in the entropy the change will take place in either direction. 

Goodenough shows further that only one part of a system 
may be considered and then the increase of unavailable energy 



J- o 7p or l ( 



, CdQ 



may be accomplished by adding heat Q or if the temperature 

T X is not constant by adding f d Q. Since this heat may come 

from internal friction, as well as from the outside, and the 
symbol Q refers to heat added, H will be used for the heat de- 
veloped by internal friction. In this case then 

Entropy change = y+ I y 

There could be a loss of availability and consequently an in- 
crease of entropy if there was internal conduction, but this need 
not be considered as in most problems the system is assumed at 
a uniform temperature throughout and any change in one part 
is the same in all parts. If S be assumed for the symbol of 
entropy 

j dS =jf + jf 

Since the amount of unavailable energy is dependent on the 
energy in a body and its temperature, the entropy, which is 
the unavailable energy divided by the lowest possible tern- 



18 HEAT ENGINEERING 

perature, is also dependent on these. Hence entropy depends 
on the state of a body and dS must be an exact differential, 
giving 

A-sx-Tf+rf as) 

J Ti UTx 

ENTROPY AROUND A CYCLE 

Now dH is the heat of internal friction and is positive in all 
cases. Hence although around a closed cycle I c^aS = 0, since a 

return is made to the original point, f -=j- must always be 

positive and therefore in such a case I -^ must be a negative 

quantity. When there is internal friction I -=f is negative 



around a closed cycle. If, however, dH = there is no friction 
and 



r- 



around a closed cycle. This friction when it exists always acts 
to produce heat no matter in what direction one progresses along 
a path. It is this which makes a theoretical path of any form 
on the pv plane an irreversible path. Hence the following 
statements are made: 

s 2 - Si = 

on any closed cycle. 

on a closed reversible cycle. 

S 2 - Si = > f^ (20) 



on a closed irreversible cycle (internal friction present), 
Now 



or 



JTdS = fdQ + jdH 



/ 



FUNDAMENTAL THERMODYNAMICS 19 

ENTROPY DIAGRAM 

TdS is an area between a curve of coordinates T and S 



and the axis of entropy and it must equal the heat added from 
the outside if dH = or the line is reversible and if dH does 
not equal zero it represents the heat added from the outside, 
plus the heat added by friction. 

With these coordinates, lines of constant temperature become 
parallel to the S axis or horizontal while lines of constant entropy 
are vertical lines. The areas beneath any line represent the 
heat added from the outside plus the internal friction. If 
there is no friction (the line is reversible), area beneath the line 
represents the heat added from the outside. If the lines are 
reversible the line of constant entropy must be an adiabatic since 



dS 



=/?=» 



and because T does not equal infinity, dQ must equal zero. This 
is the condition for an adiabatic. If the lines are adiabatic lines 
with internal friction the entropy increases due to this friction 
and hence the lines must progress to the left. 

CHARACTERISTIC EQUATIONS 

A substance of unit weight under a definite hydrostatic pressure 
and at a given temperature will occupy a certain volume and if 
the pressure and temperature change, the volume will change. 
In most cases there is for every substance a functional relation 
between these three properties or 

0(p, v, t,) = (21) 

This is called the characteristic equation of the substance 
and for each substance there is assumed to be such an equa- 
tion determined by experiment. This equation being between 
three coordinates is the equation of a surface. 

The equations for some of the well-known substances used in 
heat engines are as follows: 
For perfect gases 

pv = BT or (22) 

pV = MBT (23) 



20 HEAT ENGINEERING 

For saturated steam 

T 

log p = log k - n log T _ b (24) 

For saturated vapors 

log p = a + ba n + c/3 n (25) 

where n — {t — t Q ) 

For superheated steam 

v + c = ~- - (1 + op) J (26) 

For other superheated vapors 

pv = BT - cp n (27) 

HEAT ON A PATH 

• ♦ 

Now to study any path on which heat is added the path on 
the surface represented by the characteristic equation could be 

projected on any one of the three 
planes, pv, pt or vt. The first one of 
these is the best since area beneath 
the path represents external work but 
further than this there is no reason for 
its use. Suppose for instance the path 
is shown in Fig. 6. To find the heat the 
v path is changed to the broken path of 

Fig. 6.— Path^on the pV constant preS sure and constant volume 

lines of differential length as shown. 
The heat to change the volume by unit amount on the constant 

pressure line is ( — ) and the amount to change the pressure by 

unity on a constant volume line is (t— ) . In the broken path 

the change from a to 1 is dv and that from 1 to b is dp. These 
quantities are arbitrary in amount if the path is not fixed and 
hence they are independent variables. The heat added on a 
path is 

d * - ($,* + 0*p ^ 

It must be remembered that for a definite path only one of these 
is an independent variable since the path necessitates a relation 
between the two variables. 




FUNDAMENTAL THERMODYNAMICS 21 

In the same manner, if the path is projected on the plane pt, 
or vt, the expressions would be 

or 

If a pound of substance is referred to, these may be written 
with small letters. 

The equations are the fundamental differential equations 
of heat. 

The quantities (—J , etc., which represent the amount of 

heat added to 1 lb. of substance to change one of the proper- 
ties, p, v, or t, by unity when another property, p, v, or t, remains 
constant, are known as thermal capacities. Since these are ex- 
pressed in heat units which refer to water their values have the 
definite names given below 

(—) = c v , specific heat at constant volume. 



\8tJ „ 



specific heat at constant pressure. 
v 

These are the amounts of heat to change the temperature 

of unit weight by one degree when either the volume or pressure 

is constant. 

(/TV = ^ Vi l a ^ en ^ nea/ k °f expansion. 

This is the amount of heat added to 1 lb. of substance at 
constant temperature to change the volume by unity. It is 
latent because the temperature does not change. 
■Sq' 



\Sp)t 



l p , latent heat of pressure change. 

t 



© 



n, heat of expansion at constant pressure. 

v 

ij-J = o, heat of pressure change at constant volume. 

Using these symbols the three equations may be simplified 

dq = c v dt + l v dv (29 r ) 

dq = c p dt + l p dp (300 

dq = ndv + odp (280 



22 HEAT ENGINEERING 

It will be remembered that dq is not an exact differential 
and hence there is no differential relation between the coefficients 
of any one of the equations. There are relations which may be 
made by equating any two of the equations since these are ex- 
pressions for the same quantity. Hence 

c v dt + l v dv = Cpdt + Ipdp (31) 

This appears to be an equation with three independent vari- 
ables. There is always a characteristic equation with p, v, and 
t, and hence any one of these three is known in terms of the 
other two. Suppose t is eliminated by 



'dt\ , . (U 

\bp 
Hence 



*-£)*+©* 



[ c » (S + l -] dv + c - © f p = i Cp © , + l i dp + c ° © ,* 

Now this equation is true for any value of dv and for any value 
of dp if a definite path is not assumed, or in other words, it is true 
for any path. The only way that this can be true is to have 
the coefficients of the same variable on each side of the equation 
equal. Hence , 

c *©„ =c *©+** 



or 



and 



or 



,8p/ v p \8ps 

dp^ 



e,-C-~l,(£l (32) 

Cp C v = l v ( — ) \y<->) 



Eliminating dv from (31) leads to 

*■©.- 1 » (34) 

and 

c p - c v = l v (— J (33) 



FUNDAMENTAL THERMODYNAMICS 23 

Eliminating dp from (31) leads to 

iv ip \8v/ t 
and 

'8p^ 



c _ c = -i m 

c p c v L P\8t) 



which have been found before. By equating (28') with either 
(29') or (30') other relations could be found. 



» = c4) p + Z„ (35) 

°= c *(|)„ < 36 > 

DIFFERENTIAL EQUATIONS AND RELATIONS 

Now 

Jdq = du + pdv 

and for reversible paths (dH = 0) 

dq = Tds 

Tds 
.'. du = Jdq — pdv = — -, pdv (37) 

Another quantity used throughout the theory of thermody- 
namics which has an important use is the heat content. This is 
not the heat contained in a body but it is that heat plus the 
product of the pressure and volume. This is the amount of 
energy which would leave with a substance when it is forced 
out of a region in which the pressure is kept constant. It is 
represented by the symbol i and defined by 

i = A(u + pv) (38) 

I = Mi 
M = weight of the substance. 

It will be seen that i depends on the state of the substance 
since u, p, and v are all dependent on the state. It is expressed 
in heat units. 

di = Adu + Apdv + Avdp 
Substituting from (37) 

di = Tds + Avdp (39) 



24 HEAT ENGINEERING 

Two other quantities known as thermodynamic potentials, 

F and <f>, are given by the equations 

F = Au - Ts 

<t> = Ai - Ts 

These are both functions of the state, since u, T, s, and i are 
fixed by the state. Their differentials reduce to 

- dF = sdT + Apdv (40) 

d4> = Avdp - sdT (41) 

Now du, di, dF and dcj> are all exact differentials since they 
depend on the state and hence, as they are expressed as functions 
of two variables, the partials of the coefficients of the independent 
variables with regard to the other independent variables are 
equal. Taking the equations (37), (39), (40) and (41) the follow- 
ing results are true: 



\8v 1 s \ ds/ v 
\8p/ s \8s/ p 


(42) 
(43) 


/8_s\ _ (dm 
\8v) T \ 8tJ v 
/8s" /8v\ 
\8p) T \8t) p 


(44) 
(45) 


The four equations above are Maxwell 
equations. 
Now 


thermodynamic 


f-d, 




Hence (44) and (45) reduce to 




m- at( s 4)=l 


(46) 



Substituting (46) and (47) equations (29') and (30 r ) become 

dq = c v dt + At(^) dv (48) 

dq = c p dt - At(^) dp (49) 



FUNDAMENTAL THERMODYNAMICS 25 

Substituting (36), (35), and (33) in (28') this becomes 

dq = [*(!),+ ^ - Cj (f ) J* + c \fX dv ' 

^8p; 

'8v^ 



,8pJ 

= c »©* + c °©^ (50) 



Now 



and 



and 






(-) = 

\8v/ n 



(-) 



Op Gy 



*-zh [*©.*+ -©.*] <»> 



.<5p/ v c p — c. 
Equation (50) becomes 
AT 7 

^p Ctf l_ \ 1/1/ / t) \ L/l// p 

(48), (49) and (52) are in the same form since the partial de- 
rivatives are all derivatives with regard to dT. 

Equation (51) is an important relation which leads to valuable 
results. 

Goodenough derives two other equations which are of value. 
These follow: 

du = Jdq — pdv 
using (48) this becomes 



du 



= Jc v dt+ [t(^) - p]dv (53) 

di = Tds -f- Avdp 
= dq + Avdp 

using (49) this reduces to 

di = c p dt - A [t(^J -v]dp (54) 



26 HEAT ENGINEERING 

du and di are exact differentials. Hence 



and from (54) 



\8v/t St ^ 6t 2 8t 



These equations (55) and (56) are of value in making certain 
deductions. 

Equations (46), (47), (48), (49), (50), (51), (52), (53), (54), 
(55) and (56) are important equations in the development of 
thermodynamics. It must be remembered that all of these 
are general equations and refer to any substance. The partial 

derivatives, such as (— ) , are found from the characteristic 

equation of the substance used in any problem. 



PERFECT GASES 

A perfect gas is a substance which obeys the law of Boyle and 
the law of Charles. From these two laws the characteristic 
equation becomes 

pv = BT (57) 

or 

pV = MBT (58) 

p = pressure in pounds per square foot 
v = volume of 1 lb. of substance in cubic feet 
V = total volume of M pounds in cubic feet 
M = weight in pounds 
B = a, constant 
T = absolute temperature 
= Fahr. temp. + 459.6 



Now 



pv _ p_ 
B - T ~ mT W) 

m = weight of 1 cu. ft. of gas. 



FUNDAMENTAL THERMODYNAMICS 27 

If for air, the weight of 1 cu. ft. is found to be 0.08071 lbs. 
per cubic foot at atmospheric pressure and at 32° F. , B is found 
to be 

2116.3 

~ 0.08071 X 491.6 " 53,34 

By the law of Avogadro equal volumes of gases at the same 
pressure and temperature contain equal number of molecules 
hence, if wt = molecular weight 

m = Kwt 

• z? _ P° _ cons k 
' Kwt To " wt 

or 

Bwt = const. = R (60) 

R = universal gas constant. 

The weight of oxygen per cubic foot under atmospheric 
pressure and at 32° F. is 0.089222 lbs. 

B 2116.3 

^oxygen 0#0 89222 X 491.6 ~ 

R = 48.25 X 32 = 1544 
R - 1544 (ku 

Bgas - ^iZ (61) 



Hence 
and 



Now the molecular weight of air is found from the fact that 
air contains 79 per cent, by volume of N2 and 21 per cent, of 
O2. By Avogadro's Law, considering one molecule to each unit 
volume, the weight is equal to volume multiplied by the molecular 
weight. 

0.79 X 28.08 = 22.18 

0.21 X 32 = 6.72 



wt. of 1 vol. = 28.90 
Hence 

B *544 

air ~ 28.9 ~ b6 ' 

The correct value of this is 53.34 showing that the molecular 
weight of air is 28.93. 

If there is a mixture of various gases in a given volume there 
are two points of view to be considered: (a) all of the constituent 



28 HEAT ENGINEERING 

gases occupy the whole volume and each is under a partial 
pressure such that the sum of the pressures equals the total 
pressure, or (b) each gas is under the total pressure but occupies 
a part of the volume so that the sum of the partial volumes 
equals the total volume. The first view is known as Dalton's 
Law and represents what really takes place when a number of 
gases are mechanically mixed. If pi, p 2 , Ps . . . represent the 
partial pressures of the various gases which weigh M h M 2 , M 3 
. . . and if all the gases occupy the volume V the following 
are true: 

p = pi + p 2 + ?>3 • • • (Dalton's Law) 
M = M i + M 2 + M 3 . . . 

p = M 1 B 1 ^ + M 2 B,^+ . . . = |sMJ5 

V 

Py= MB mixture = VM 1 B 1 

or 

■D mixture y lur \"^J 

This is not as convenient a method as working with volumes 
since in most cases it is the proportional volumes which are 
known v and not the proportional weights. By Avogadro's Law 
each unit of volume may be assumed to have one molecule. 

V\wti = proportional weight. 
'ZViwti = S proportional weights = total weight 

If this is divided by the sum of the volumes, the weight of 
unit volume, which by assumption is the molecular weight, is 
found. 
Hence 

ZViwh _ 

1544 

■^mixture = ~~> \04J 

wl mixture 

In the above manner by (61), (63) and (64) the value of B 
for any gas or mixture may be found if the gas is assumed to be a 
perfect gas. No gas obeys the laws of Boyle and Charles com- 
pletely and hence the characteristic equation 

pV = MBT 
is not exactly true for any gas. Under great pressures there is 
variation, but air, hydrogen, nitrogen, carbon monoxide, meth- 



FUNDAMENTAL THERMODYNAMICS 29 

ane, ethylene, and other hydrocarbons are assumed to follow 
these laws. Carbon dioxide and steam do not obey the equa- 
tion, but at times they are handled as if they did. 

To aid in working out the values of B the following molecular 
weights are given: 



Hydrogen 


H 2 


2.016 


Carbon monoxide 


CO 


28.0 


Oxygen 


o 2 


32.0 


Methane 


CH 4 


16.032 


Ethylene 


C2H4 


28.032 


Nitrogen 


N 2 


28.08 


Carbon 


C 


12.0 


Ammonia 


NH 3 


17.064 


Carbon dioxide 


C0 2 


44.0 


Steam 


H 2 


18.016 


Sulphur 


S 


32.06 



Sp\ 



DIFFERENTIAL HEAT EQUATIONS FOR PERFECT GASES 

For a perfect gas 

pv = MBT 
or 

pv = BT 
and the various equations (46), etc., are reduced by the following: 

B 

V 

'Sv\ = B 

8t ) P ~P 
It is to be remembered that although t + 459.6 = T 

St = ST. 

U = ^~ (46) 

ATB 

dq = c v dt + dv = c v dt + Apdv (48) 

ATB 

dq = Cpdt dp = c p dt — Avdp (49) 

dq — c p -^ dv +c v ^dp = c p T h c v T — (50) 

c p -c v = AT- - = AB (51) 

p v 



30 HEAT ENGINEERING 

In equation (48) the last term represents the external work, 
therefore the first term represents the change in intrinsic energy : 

Jc v dt = du (65) 

Now du is an exact differential and must be directly integrable. 
Hence c v dt must also be directly integrable (i.e., with no reference 
to any path) and therefrom c v for a perfect gas is a constant 
or a function of t. 

SPECIFIC HEATS 

Now experiment proves that c v is a function of * of the form 

c v = a + bt 



or 



But 



c v = a + b(T - 459.6) = a - 459.66 + bT 

= a' + bT (66) 



/ qy t/'U "~~ .ijL J-J 



c p = AB + a' + bT = a" + bT (67) 



Now since 



Cp ~~ Cy Ji-IJ 

c p , c v , and A B must be of the same nature, if c p and c v are 
specific quantities AB must be a specific quantity. Since c p 
is the amount of heat to change the temperature one degree and 
since c v dt = du and therefore c v is the amount of internal energy 
change when the temperature changes one degree on any line, 
AB must be the amount of external work when the temperature 
changes one degree at constant pressure. 
Since 

wt X B = R, a constant 

wt X c v = a xn + 6 111 T 

wtXc p = a IY + b™ T 

are the universal forms of specific heats and these differ by AR. 

1544 

wt X c p — wt X c v = AR = -==£■ = 1.9855 



The values of the a's and b's are given below 
wt X c v = 4.77 + 0.000667* ' 
wt X Cp = 6.75 + 0.000667* 
For CO 2 

c v = 0.15 + 0.000066* 
Cp = 0.195 + 0.000066* 



i 

| for any perfect gas. 



FUNDAMENTAL THERMODYNAMICS 31 

For superheated steam. (Approximate.) 

c v = 0.324 + 0.000133* 
c p = 0.435 + 0.000133* 
Although 

Cp — C v == JiJlj 

is always true the relation 

^ = k = 1.4 (68) 



is approximately true for ordinary temperatures. Equations 
(51) and (68) are not both possible for two such equations could 
only hold for definite values of c p and c v . (68) is approximately 
true since the value of the quantity b is small and as it gives 
simpler results to certain problems its use is advisable unless 
there is a great change in temperature on the path considered. 
Since 

Cp — C v — J\.£j 

and Cp = kc v 

AB = (k- IK 
or Cp : c v : AB = k : 1 : k — 1 

ISOTHERMAL AND ISODYNAMIC LINES 

If in equation (65), T is made constant du will be equal to 
zero, or u is constant. Hence for perfect gases the isothermal 
is the same as the isodynamic. 
Since 

pv = BT 

pv = constant 

is the equation of this line on the pv plane. This is the equation 
of the rectangular hyperbola. 

ADIABATIC 

The adiabatic is such a line that dq = 0. 
From (48). 

= c v dt + Apdv 

c v dt = — Apdv = — A — dv 

c v dt^ _ dv 
ABY~ ~~v 
. c v T 2 . vi 

..^loge^loge- 



32 HEAT ENGINEERING 

or 

T 1 ~ \v 2 



\w 



but 

(-"0 

hence 



a-er' 



or 

rp v k-l = cons t # 

This is the equation of the adiabatic on the v T plane. To 
reduce this to the equation for the pv plane, T must be eliminated. 
Now 









T = 


B 








Hence 






pv v k_1 
B 


const. 








or 






pv k = 


const. 






(70) 


This is 


the 


equation of the adiabatic of a perfect 


gas 


on the 


pv plane. 
















Now 






v = 


BT 

V 








Substituting this in 


(70) gives 
















»(t)'- 

k-1 


const. 









p k T = const. 



(71) 



(69), (70) and (71) are the projections of the adiabatic line 
of the surface pv = BT (perfect gas) on the three planes of 
projection. 

The line pv n = const, is known as a polytropic and for the per- 
fect gas this has the three forms on the different planes: 

pv n = const. (72) 

y — T = const> (73) 

v n-irp = const (74) 



FUNDAMENTAL THERMODYNAMICS 



33 



THERMODYNAMIC LINES 

The six important lines used in thermodynamics are the 
isothermal (constant T) , adiabatic (no heat added from the out- 
side), isodynamic (constant intrinsic energy), constant pressure, 
constant volume and polytropic. 

For a perfect gas these lines on the pv plane have the following 
equations: 

Isothermal, pv = constant 

Adiabatic, pv k = constant 

Isodynamic, pv = constant 

Gonstant pressure, p = constant 

Constant volume, v = constant 

Polytropic, pv n = constant 



n = lA 



w = l 



71 - Positive «, 
n = \ 


n 


= 00 

n, Negative 

P = Const, n = 


n -"Negative 


V 

n 


s ^. n -Positive 

= Const. ^sf° A "^^ 
- 00 x- 



Fig. 7. — Thermal lines. 



It is seen that each is of the form pv 1 
n has certain values. 



constant in which 



n = 1 for isothermal and isodynamic 

n = k for adiabatic 

n = for constant pressure 

n = oo for constant volume 



These are shown in Fig. 7. 

Since these are all of the same form it will be well to investigate 
the general case first. 

3 



34 HEAT ENGINEERING 

pv n = const. 
dq = c v dt + Apdv 
du = Jc v dt 
dW = pdv 
AB 
Cv ~ k - 1 

.-. ( U2 - Ul ) = Jc v (T 2 - T,) = j^-j (T 2 - TO = 

dw = I pdv = const. I v~ n dv = const, y- — — 

t/ VI t/ vi -'n 

const. 



(75) 



— V - v n 




Now Const. = piVi n = p 2 v 2 n 




W— U P2^2^2 1_n - PlVi n Vl 1_n p 2 l>2 " 


- Pi^i 


W OI*K — ., — _, 

1 — n 1 - 


- n 


When w = 1 




Z>2^2 = PlVl 




and the expression for work reduces to 




Work = jj. 





an indeterminate expression. To find the value of the work in 
this case, the integration must be made by using the substitution 
for this case. 

Work = I pdv = const. I — = const. log e — = pit>ilog e — (76) 

X X v Vi "i 

since const. 

p = 

^ v 

The value (75) for work holds for all values of n except for 
n = 1. In this case (76) is the expression for work. Equation 
(75) may be put in two other forms, thus: 

Work . m^m = LA — &»L = \ W / (77) 

1 — 71 1 — 71 1 — 71 

= \ X -^—L (770 

1 — 71 



FUNDAMENTAL THERMODYNAMICS 35 

Since 



£>2t>2 ?>2 \??2 



©-• 



or 



work = LVp/ J_ LV J 

1 — n 1 — n v ' 

The expressions for the heat become: 

= pm - m v** ~ Pi* (79) 

* h — 1 1 — n v y 

for all values of n except n = 1, and 

q = PlV! loge ^ (80) 

for n = 1 since there is no change in intrinsic energy on the 
isothermal. If these do not refer to 1 lb. but to M lbs. 

C-k.-^j^ + rr-J (790 

(800 



V, 



Q = PlVllOgeY 

It will be seen from (790 that 

+ ■: is proportional to the heat, 



k — 1 1 — n 
, _ .. is proportional to the change of energy, and 
1 



1 — n 



is proportional to the work 



along the polytropic pv n = const, of a perfect gas. It must be 
remembered that this is only true for perfect gases and on 
polytropics. 

For n = k, the expression (79) becomes 

?>2^2 — PlVl PlVl — P\V\ _ n 

ft- 1 + 1-fc " U 

or the intrinsic energy change 2 jf _ ^ ] is equal to minus the 

work. If the final volume is infinity this in the form of (78) 
reduces to 

Work = ^ 



36 



HEAT ENGINEERING 



This is the work under an adiabatic to infinity from the point 
1 and hence it is the expression for the intrinsic energy at the 
point 1. 

The expressions (75) and (76) for work are expressions for 
areas beneath lines and consequently the expressions are true 
for all substances expanding on these paths. 

Since p 2 T^2 = pi^i on the polytropic n = 1 there is no change 
in intrinsic energy on this line for a perfect gas and it is the 
division line between positive and negative energy changes. 

On the line n = k there is no heat added and so this is the 
division line between positive and negative heat. 




Fig. 8. — Lines of division between positive and negative quantities. 

The constant volume line is the division line between positive 
and negative work. 

These are shown in Fig. 8. 



EQUALITY OF GAS SCALE AND KELVIN SCALE 



Now 



pv = BT 
has been derived from the definition of the perfect gas from which 

(81) 

From (53) which has been derived from absolute considerations 

du = Jc v dt + ]t ^ - pjdv 



*P _ B _ P 

bt ~ v ~ T 



FUNDAMENTAL THERMODYNAMICS 37 

Since du is independent of v for a perfect gas by the experi- 
ments of Joule and Thompson: 

(Joules' Law) T (^j -p = 

(81) and (82) give the same result, hence the absolute Kelvin 
T of (82) must be the same as the perfect gas T of (81). 



HEAT CONTENT OF GASES 



Since u = 



k- 1 
i=A[^- 1 + V v]=A^ lV v (83) 

For a change 

l2-Ii=A j^rj(v2V 2 - Pl 7i) (85) 

ENTROPY OF GASES 

For reversible lines 







ds = 


dq 

T 


hence from (48) 


. (49), 


and (50) : 




s 2 — Si = I c v 




(a \dv 


= 



- si = c p log e ^ — AB log e P (87) 



r; - AB log < * 



s 2 — si = c p log e — + c v loge — (88) 

These changes in entropy depend only on the states at the 
beginning and the end of the path. 



CYCLES AND CROSS PRODUCTS 

Cycles are the paths showing the changes in the properties of a 
substance as it undergoes a change. They are often shown on 
the pv plane as in Fig. 9. A simple cycle is one made up of two 



38 



HEAT ENGINEERING 



pairs of similar lines. A cycle of four or more different lines is 
a complex cycle. 

The cycles of engines using perfect gases have certain properties 
provided that they are made up of two pairs of polytropics. If 
the simple cycle is made up of polytropics, pv n = const., 

(89) 
(90) 
(91) 



V 1 V 3 


= V 2 V 4 




ViPs 


= P2PI 




T^s 


= T 2 T, 




i. 






\ >. pv n 


= Const . 




\ pv m = 


■ Const\ 






\ pv m ^ 


Const. 


4* 








P^C^t 


*3 



Fig. 9. — Simple cycle of polytropics. 
This is shown as follows: 

PlVi n = p 2 V2 U 
p 2 V 2 m = P3V 3 m 
P 3 V 3 n = p 4 V4 n 

p 4 v 4 m = p^x™ 

PlP2P3P4(VlV 3 ) n (v 2 V4) m = PlP2P3P4(VlV3) m (v 2 Vi) n 

(viv 3 ) n - m = (v 2 v 4 ) n - m 

ViV 3 = V 2 V± 

11 11 



(p lP3 )n m= (p 2Pi )n 
P1P3 = P2P4 



(89) 



(90) 



If the cross products for pressures and for volumes are equal, 
those for temperature must be equal. 



B*T x Tz = B 2 T 2 T 4 
T X T 3 = T 2 T, 



(91) 



This latter is true for perfect gases only. For cycles of any 
substance the cross products of pressures or volumes are equal 



FUNDAMENTAL THERMODYNAMICS 39 

from the geometry of the figure but the cross products of tem- 
peratures are only true for perfect gases. 

SATURATED STEAM AND OTHER VAPORS 

A vapor is a gaseous condition of a substance near its point of 
liquefaction. 

When a vapor is in contact with its liquid it is said to be 
saturated. 

The characteristic equation for steam (due to Bertrand) is 

T 

log p = log k - n log T _ b (92) 

p = pressure in pounds per sq. in. 
T = absolute temperature in deg. F. 
n = 50 

32° F. to 90° F. 90° F. to 237° F. 238° F. to 420° F. 

b = 140.1 b = 141.43 b = 140.8 

log k = 6.23167 log k = 6.30217 log k = 6.27756 

For other vapors Goodenough gives the Dupre-Hertz formula 
with the value of constants from Bertrand. 

C 



logp 


= a - b log T - j, 




(93; 


V 


= pressure in millimeters of mercury 




T 


= absolute temperature 


in degree C. 




Water 


a 
17.44324 


b 
3 . 8682 


c 
2795.0 


Ether 


13.42311 


1.9787 


1729.97 


Alcohol 


21.44687 


4.2248 


2734.8 


Sulphur dioxide 


16.99036 


3.2198 


1604.8 


Ammonia 


13.37156 


1.8726 


1449.8 


Carbon dioxide 


6.41443 


-0.4186 


819.77 



These equations are rarely used since tables of the properties 
of vapors have been constructed giving the pressures and corre- 
sponding temperatures. From the characteristic equations of 
saturated vapors it is seen that the pressure and temperature 
are independent of the volume. When heat is added to 1 lb. 
of liquid at 32° F. and of volume v' it is found that the volume in- 
creases a very slight amount so that dv may be considered as 
zero, giving dq = c v dt, or better cdt to include the slight amount 
of work. 



40 HEAT ENGINEERING 

The value of c has been determined experimentally for dif- 
ferent temperatures, hence if these be plotted as functions of t, 
the area beneath the curve will give the value of the integral 






cdt (94) 



This is called the heat of the liquid. It is the amount of heat 
to raise 1 lb. of liquid from 32 to some temperature t. For 
water it is found graphically by plotting c but for other sub- 
stances the quantity is found by empirical equations of the form 

q' = a + bt + ct 2 + . . . (95) 

c = ^L = b + 2ct + Sdt 2 + . . . (96) 

After the liquid has been heated to the temperature corre- 
sponding to the pressure, the addition of heat causes the liquid 
to boil and some of the liquid is changed into vapor. When 
all is changed into vapor the volume of 1 lb. is v" so that 
the volume has been changed by the amount 

v" — v' = change of volume. 

If x represents the amount of 1 lb. which has been changed 
into steam it is called the quality, (1 — x) is the amount which re- 
mains liquid. The volume of the 1 lb. of mixture is 

v = v ' + x (v" - v') = (1 - x)v' + xv" (97) 

V = M [(1 - x)v' + xv"] (98) 

Since v' is small and since (1 — x) is small in most cases, 
(1 — x)v f may be neglected, giving 

V = Mxv" (99) 

The amount of heat added to change 1 lb. of liquid from 
liquid at the boiling point to vapor at this temperature is called 
the heat of vaporization and is represented by r. 

r = j dq = ( c v dt + J AT (^J dv 
Now T is constant and ( — ) is independent of v, hence 

r - A K«)jT* 

= AT Q) y - v') (100) 



FUNDAMENTAL THERMODYNAMICS 41 

This equation may be used to find r if v" — v f is known. 
Since r is usually found by an empirical equation, equation (100) 
is used to compute (v" — v'). 

For steam 

r = 970.4 - 0.6550 - 212) - 0.000450 - 212) 2 (101) 

The total heat added to 1 lb. of liquid at 32° F. to make it 
into steam at a given temperature is called the total heat, q" or 

q" = q' +r (102) 

For steam 

q" = 1150.4 + 0.350 - 212) + 0.000333(t - 212) 2 (103) 

For other substances the equations may be found in tables of 
properties. 

Having r, v" — v' may be found; since by (92), 

dp _ nbp 
dt ~ T(T - b) 

v" - v' = — 



nbp (104) 

A T{T -b) 

If the volume is changed by v" — v' at a pressure p, the 
external work expressed in heat units is 

A work = Ap{v" - v') = ^ (105) 

The internal heat of vaporization is therefore 

p = r - V (106) 

This is computed and placed in the tables. 

If the steam or vapor is such that no liquid is present, x = 1 
and the vapor is dry. However, if there is some liquid, x is not 
unity and the vapor is wet. In either case it is a saturated vapor 
as the pressure and temperature are related by means of the 
characteristic equation of saturated vapor. When x is not 
unity the total heat is 

qj' = q' + xt 
INTRINSIC ENERGY 

There is practically no work when the liquid is heated, hence 
the quantity of heat q' remains in the body and when steam is 
made the change of intrinsic energy is given by 

A(u — u d2 ) = q' + r — Ap(v" — v') 
= <Z' + P 



42 



HEAT ENGINEERING 



Since the intrinsic energy may be measured from 32° F., this 
may be written 

An = q' + P , or AU = Af (g' + p) (107) 

For wet vapor 

Au = q' + zp, or AC/ = Af (g' + xp). (107') 

HEAT ON A PATH 

If heat is added to a vapor on some path as in Fig. 10 

q= A(U 2 - *7i) +A CpdV 
= AM [q' 2 + X2P2 - q'i - x lPl ] + A ( pd7 



Fig. 10. — Path on pF plane. 

The area beneath the curve is the value of the integral. This 
may be of any of the forms given earlier. 
To find x the formula 

V 



x = 



Mi/ 



(108) 



is used. 

HEAT CONTENT 

At times steam tables and charts give the value of the heat 
content and the method of using this must be clearly understood. 

i = A (u + pv") 
= q' + r' - Ap(v" - v') + Apv" 
= q f + r f — Apv\ 

Now Apv' is a very small quantity, hence i and q" are almost 
the same. If, however, i is given 

i - Apv = Au (109) 

so that for A(u 2 — U\) 

i 2 — Ap 2 v 2 — (t! — Aptfi) 

may be written and problems solved. 



FUNDAMENTAL THERMODYNAMICS 43 

ENTROPY 
The entropy change when the liquid is heated is 

S t - S 32 = I c - = I cd(log e T) = s' (110) 

^/491.6 J- J 91.6 

This entropy of the liquid is found graphically as the area 
under a curve of c plotted against log T. 

When the liquid at the boiling point is vaporized 
f x = 1 f'dq Crdx r f 1 r 

X d Jr)T=J^F=TV x =T (m) 

The total change of entropy from liquid at 32° F. into dry 
vapor is 



T 
S" - S 3 2 = S' + p 




Since 32° F. is the datum plane of reference 




s" = 8' + £ 


(112) 


XT 

For wet vapor s" = s' + -= 


(1120 



DIFFERENTIAL EFFECT OF HEAT 

When a mixture of liquid and vapor has a differential amount 

of heat added to it, an amount dx of liquid is vaporized, and the 

liquid (1 — x) and the vapor x are raised dt degrees. If c' is 

the specific heat of the liquid and c" that of the vapor, 

dq = c'(l - x) dt + c" X dt + rdx 

, T , dq c'(l - x) + c"a; „ , r 7 
Now ds = -~ = ™ dt + -ydx 



This is an exact differential, hence 

x) + c r/ x \ _5_ /_r \ 

T 7 / r ~ STAT 7 / 



_5_ /</(l - *) + c ,r ^ 

T^~ T ~ T bT T 2 



-e'-'J + S (113) 



c =df 
and dq^_<W <fr 



C - dT T {ll6) 



44 HEAT ENGINEERING 

Since q" = a + bit - 212) - c(t - 212) 2 

^ = b - 2c(Z - 212) 

may be substituted and c" may be found. 

SUPERHEATED VAPOR 

If the saturated vapor is taken away from its liquid so that 
additional heat will not vaporize any liquid, the addition of 
heat will cause the temperature to rise above its saturation 
value if the pressure is kept constant. This vapor is known as 
superheated vapor. 

The equation for superheated steam as reduced by Good- 
enough from the work of Knoblauch, Linde and Klebe is 

v + c = ^ - (1 + ap) Yn (H4) 

p = pressure in pounds per square inch 
v = volume of 1 lb. in cu. ft. 
B = 0.5963 
log m = 13.67938 
n = 5 
c = 0.088 
a = 0.0006. 

For other substances the equation of Zeuner is used 

pv = BT - cp n (115) 

For high temperatures Mallard and Le Chatelier and Langen 
give for the specific heat of superheated steam at constant 
pressure 

c p = 0.439 + 0.000239* (116) 

but from (56) according to Goodenough 

(11= -«s 

From (114) 8v B ' mn . 

8*v_ _ mw(w+.l) __ (8cA _1_ 

8t 2 ~ ~ Tn + 2 U + ap; - \ Sp ) T AT 

Amn(n — 1) / a \ _ ni ->. 

.-c p = — ^+i — p[l + zP) + 4>T (117) 



FUNDAMENTAL THERMODYNAMICS 45 

Using (116) as the form of expression for <f>T 

nrr , , Amnin — 1) /., , a \ 
c p = a + (3T + j^+j J -p(l + -p) 

From the results of Knoblauch and Mollier, Goodenough re- 
duces for a and /3 the values 

a = 0.367 
= 0.0001 

giving c p = 0.367 + 0.00017 7 + p (1 + 0.0003p) ^ (1170 

log C = 14.42408 

p = pounds per square inch 
T = degrees absolute F. 

This value of c p agrees with the results of experiment by 
Knoblauch and Mollier. Values of this are shown in Fig. 11 as 
given by Goodenough. 

The value of the specific heat of other superheated vapors 
is given below and these are usually taken as constants although 
this is not true. 

SPECIFIC HEATS AT CONSTANT PRESSURE AND K'S 



Superheated ammonia vapor 


0.536 


k = 


1.32 


Superheated sulphur dioxide 


0.1544 


k = 


1.26 


Superheated carbon dioxide 


0.215 


k = 


1.30 


Superheated chloroform 


0.144 


k = 


1.10 


Superheated ether 


0.462 


k = 


1.03 



If now 1 lb. of liquid at 32° F. is heated and finally changed 
into superheated vapor at temperature T SUP . the amount of heat 
required is given by 

r = 2' + M \c p dt (118) 



f*T eup . 

I c p dt 

J Ttat. 



The last term is found graphically or analytically for different 
pressures and different amounts of superheat and tabulated or 
plotted in charts so that q'" may be known. 

The value of I c p dv may be written as 

J Teal. 

(mean c p ) (T 8up . - T 8at ) 



HEAT ENGINEERING 




200 300 

Degrees of Superheat 
Fig. 11. — Specific heat of superheated steam. 



500 



FUNDAMENTAL THERMODYNAMICS 47 

The value of mean c p is given by 



mean c, 



I c p dt 



J- sup. J- sat. 

The values of mean c p have been compared and plotted in the 
form of curves by Goodenough. From these curves the table 
on p. 48 has been constructed: 

The volume of this superheated steam is computed by (114) 
and tabulated or plotted. 

The heat remaining is 

Au = q"' - Ap(v'" - v') 
But Au = i - Apv'". (109) 
.-. i = q '" + Apv' 
i = q'" practically. 

The entropy of the superheated vapor measured from liquid 
at 32° F. is 

>T SU 



r-dt 

h f 

t/ T aa t. 



This last term is found analytically or graphically, but in most 
cases these have been tabulated for steam so that values of 
s'" may be found for definite conditions. 

The various quantities for saturated steam mixtures and super- 
heated steam may be found and these properties are often made 
into charts. The quantities p, v, t, s, u, i, x are all dependent on 
the state and, if known, fix the state. Usually any two of them 
fix the state, so that these two could be used for the coordinates 
of a diagram. As has been shown before, T and S could be used, 
and in this the area under a line represents the heat added from 
the outside if the line is reversible or the heat added from the 
outside together with friction if the line is non-reversible. 

Other coordinates such as i and s have been proposed by 
Mollier. This is known as a Mollier diagram. These will now 
be explained. 

T-S AND I-S CHARTS 

In Fig. 12 the coordinates T-S are used and since heat is added 
to the water at 32° F. the line of 



J491.6 



48 



HEAT ENGINEERING 



o 

8 


moco 

■>* "#CO 
CO CO CO 


t»cooo 

(NCM-H 

CO CO CD 


--HOOOO 
CO co CD CD m 


(NCOrHCOlN 

D00 00NN 
lO ID »D ID ID 


00^O»D 
CO CO ID ID 
ID ID ID m 


wooocc 

ID ID ID ID 


eco 

«DiO 


coco 
mm 


ooo 


OOO 


OOOOO 


OOOOO 1 oooo 


OOOO 


oo 1 od 


o 

m 

(M 


OC0l> 
CO CD CD 


<NI>C0 

^OO 

CO CO CD 


OlOrHOOkO 

OOOOO 00 
lO >D 10 lO ID 


00 CO 00 "CH>-I 
!>•>>. CO CO CO 
lO ID ID >D iD 


OOtJHOCO 
iD»DiD-^ 
iDiOmm 


COr-HOSt^ 

■^ rfi co co 

ID iD to tO 


TJ<(N 

coco 

tOto 


oo 

COIN 

mm 


OOO 


OOO 


OOOOO 


OOOOO 


OOOO 


OOOO 


oo 


oo 


O 

o 

IN 


OCON 
i-HOO 
CD com 


ThOCO 

ooo oo 

»OkDiD 


CM CO »D COO 

oot^t^r^t> 

ID >D >D m iC 


lOOCONO 
CO CO ID ID -^ 
ID »D ID ID »D 


m> IN 00 ID 
tF-#COCO 
ID iD >D ID 


co—tooo 

CO CO CO <N 
lO ID >D ID 


CO-* 
NCN 

ID ID 


coco 

<N(N 

mm 


OOO I OOO 


OOOOO 


OOOOO 


OOOO 


OOOO 


OO 


oo 


1 


00 coo 

OOOOt^ 
lOiDiO 


ID--I1> 
t^t^CO 

inmm> 


-* .-100 CO CO 
CO CO ID >D ID 
ID ID "D »D >D 


00COOt>"# 

^n ^ ^ co co 

lO ID ID ID ID 


CNOCDCO 
CO<N<N<N 
ID kD ID ID 


CNOO00 COiD 
NIMHH | y-lT-t 
IDiDiDiD >D ID 


mm 


OOO 


OOO 


OOOOO 


OOOOO 


OOOO 


OOOO 1 do 


oo 


HO 

<N 


INOO^ | (NOOCO 
t- CO CO 1 COiO ID 
lOiDiO 1 ift>D»D 


C0O00>D CO 
iDO-*-*-^ 
ID <D ID ID »D 


00>D<NOt> 
COC0CO<NtN 
ID iO iD ID m 


CO tH 00 CO 
(NINHH 
ID ID ID >D 


■^COCOCN 
iD ID "Dm 


OO 

HO 

ID ID 


oo 
oo 
mm 


OOO OOO 


OOOOO 


OOOOO 


OOOO 


OOOO 


OO 


oo 


O 

o 


OlOCM 1 OCOCO 

>D »D »D 1 Tt< -* -^ 

lO lOiO I iO ID iD 


000 CO CO rH 

-tfcocococo 

ID ID ID ID ID 


t>T}H(MOJ> 

(N (N CN <-H i-i 
ID >D ID >D ID 


iDCOOOO 

HHHO 

u l ID »D ID 


I>CDiD«D 
OOOO 
ID ID ID ID 


ID iD 


oo 
mm 


OOO OOO 


OOOOO 


OOOOO 


OOOO 


OOOO 


OO 


oo 




00*OCN 1 OOIOCO 
■^ Tf< TjH CO coco 
lO lO iO lO id m> 


O00 CO-* CO 
C0<N<NiN<N 

id m m> m m 


DNW5HO 
ID >D >D >D >D 


00 CO t* CO 
OOOO 
ID ID ID ID 


OOOOO 
OOOO 

IDTt*'*'* 


00 00 

oo 


00 00 

oo 


OOO 1 OOO 

1 


OOOOO 


OOOOO 


OOOO 


OOOO 1 oo 


oo 


o 

m 


<NOI> 
<Ni-HrH 


-cHlNO 
ID ID lO 


00I>iO-*c0 
OOOOO 
iDiO »D lO ID 


O00 1> ID CO 
OOOOO 
ID -<frl •<* Tf Tfl 


CNOO00 
OOOO 00 
^cH "^ "^ ^ 


00 00t^l> 
OOOOOOOO 

^ ^ T^ ^ 


00 00 


00 00 

oo 


OOO 


OOO 1 OOOOO 


OOOOO 


OOOO 


OOOO 


oo 


oo 


in 

CO 


I>iOCO 
OOO 


i-HOl> 

OOO 


IDCOCN'-IO 
OOOOO 


00 CD »D-* CO 

oooooooooo 


oooooooo 

tH ^ ^ ^ 


^HiNCNCN 

oooooooo 

^ ^ ^ ^ 


CO-* 
00 00 


00 00 


OOO 


OOO 


OOOOO 


OOOOO 


OOOO 


OOOO 


oo 


oo 


IN 


C0--IO 


oot^co 

00 00 00 


00 00 00 00 00 


ooioonco 


tDiD-^TjH 

"^ ^ "^ ^ 


Ttl-^IOIO 
"<^ Tfi T^i T^ 


COl^ 


OOO 


OOO 


OOO 


OOOOO 1 OOOOO 


OOOO 


OOOO 


oo 


oo 


lO 


oo com 


CO CON 


OOOOO 00 

i> rococo co 


i>i>cococo 

CO CO CO CO CO 


ID ID to CD 
CO CO CO CO 
^ ^ ^ tJ* 


CDt^t^OO 
CO CO CO CO 
^ ^ ^ ^ 


OOO 
COl> 




OOO 


OOO 


OOOOO 


OOOOO 


OOOO 


OOOO 


oo 


OO 


o 


COiO-^ 

CD CO CO 


CO COIN 
CO CO CO 


(Ni-HOOO 
CO CO CO CO m 


000 00 00 00 
ID ID tO to ID 


OOOOO 
iD ID ID CD 
"^ ^ ^ ^ 


.-HtNCNCO 

CO CO CO CO 

^ ^ "^ ^ 


TjHCO 

COCO 


OOO 

coi> 


OOO 


OOO 


OOOOO 


OOOOO 


OOOO 


OOOO 


oo 


oo 


t^ 


OOO CO 
iO»DiO 


l>cOCO 
to »C LO 


CD ID iD >D Tt< 

ID >D »D >D ID 


ID ID ID >D ID 


ID>D>0 CO 
ID ID ID >D 

^ Tj< ^ ^ 


l>t>00O 
ID ID id ID 
^cH ^ ^ ^ 


T-HCO 
COCO 


"AN 

coco 


OOO 


OOO 


OOOOO 


OOOOO 


OOOO 


OOOO 


oo 


OO 


iC 


CO CO CM 

mmm 


<N<-t--H 
iD»OiD 


OOOOO 
ID ID ID tP Tti 


OOOOO 

T)H T*l TfllD ID 


Ot-hCN(N 
lOiOiOiO 

^ ^ ^1 Tf 


C0-* >D CD 
iO iD ID iD 


OOO 
>n co 


wm 

coco 


OOO 


OOO 


OOOOO 


OOOOO 


OOOO 


dodo 


do i oo 


co 


^^^ 


^ tJ^ tJ< 


"^ "^ CO CO CO 


COCO-^iDO 


iOCOC01> 
^ Tt^ ^ ^ 


OOOOrH 

•^■^•DiD 


coco 
mm 


00»-l 

mco 


OOO 


OOO 


OOOOO 


OOOOO 


<6oo<6 


ooc><6 


oo 


oo 


IN 


OOO 

co co co 


SiS 


o o o o o 

CO CO CO CO CO 


OOOhin 


(NCOTjfiD 

^ ^ Tt^ ^ 
^f Tf ^H ^ 


CDI>00O 

^f ^4 T^H T^l 
^ ^ ^ ^ 


-HCO 

mm 

Tf4Tj< 


CO 00 

mm 


OOO 


OOO 


OOOOO 


OOOOO 


OOOO 


OOOO 


oo 


OO 


- 


CO CO CO 


IN CO CO 

co co co 


CO CO CO ^ *^ 

co co co co co 


iD»DcOCOl^ 1 OOfflOH 
cococococo coco^Tji 


(N CO-* iD 
^ ^i ^ ^ 


OOO 

Tjnm 

TtH'CH 


cNm 

mm 


OOO 


OOO 


OOOOO 


OOOOO 1 OOOCO 


OOOO 


oo 


oo 


-jadns jo 

S99J38Q 


OOO 


OOO 


OOOOO 
COt>COOO 

1-1 


OOOOO 
(N-* CO OOO 
.-H tH T-l 1-1 IN 


OOOO 
<N-*t^O 
<N IN <N CO 


iDOiOO 
iNiDr»0 
CO CO CO Tj< 


mo 
■*m 


mo 
mo 



FUNDAMENTAL THERMODYNAMICS 



49 



starts at 32° F. or 491.6° abs. F. This line is shown as the liquid 
line in Fig. 12. The various values of s' for different values of T 
from the steam tables are plotted. From points on the liquid 

line the distances ™ for those temperatures are laid off giving the 

saturation line. It is evident that these two lines approach 




p — « r ~""^ s 

Fig. 12. — TS diagram for TS analyses. 



Fig. 13. — TS diagram areas. 



and meet at the critical temperature (at which the superheated, 
saturated and liquid states coincide), since at this point r = 0. 



Since 
and 



dq = Tds 
dq = cdt 
,ds 



c=T 



dt 



(120) 



or the specific heat is the subtangent from any point on a curve 
on the T-S plane. In Fig. 12 it will be seen that while the specific 
heat of the liquid is positive, the specific heat of saturated steam 
is negative. 

The areas beneath lines on which there is no friction are 
equal to the heat added. As shown in Fig. 13, the area beneath 



50 



HEAT ENGINEERING 



the liquid line is q' while that beneath the line from the liquid line 
to the saturation line is r. It must be remembered distinctly 
that steam tables and diagrams assume that steam is made 
by keeping the pressure constant, heating the water from 32° F. 
to the boiling point, boiling to dry vapor and then superheating at 
constant pressure. If superheating takes place the temperature 
will rise by the degrees of superheat and the entropy will increase 

by 



J 



Cpdt 



the line will therefore take the shape cd. The broken line abed 
is a line of constant pressure. The area beneath cd is 



/■ 



c v dt 



If the steam is wet the change of entropy from the liquid 

XT 

condition is ^- , and hence if point e is so selected that 



be 
be 



= x 



the point e represents the condition of quality x since be = ^. 

If now for various temperatures (and consequently saturation 

pressures) various points be 
found, having the same 
value of x, a line of constant 
quality or constant steam 
weight may be obtained as 
shown in Fig. 14. 

In the superheated region 
this is replaced by lines of 
<* constant amount of super- 
heat. In this case points 
on various pressure lines at 
the same number of de- 
grees above saturation are 
connected. 




Lines on TS plane. 



If the value of i be computed for point e and this is made equal 
to the i at some other temperature the quality at this point may 
be found and from it the position of e on that line. 

q' + xr = q\ + xiTi 

q' + xt - q\ 
ri 



Xi = 



FUNDAMENTAL THERMODYNAMICS 



51 



If a number of points are found, a line of constant heat content 
is obtained. In the superheated region 



q' + r + 



I Cpdt = q\ + ri + j Cpdt 



and the values of the degrees of superheat at the second pres- 
sures are found to give the same heat content as at the first point. 
Such points form the line of constant heat content. 
It is known that 

u = q' + xp or u = i — Apv 

Hence if the value of u at point e is equated to the expression at 
another pressure or temperature, the quality x or the degrees of 
superheat may be found for the second pressure or temperature 
and this fixes the position of the point. Connecting a series 
of these the line of constant intrinsic energy is found. 
If the volume of the point e is found as 

v = xv" 

and equated to x x v" x for a different pressure the value of x may 
be found, and from a series of these points a line of constant 



1400 




Fig. 15. — Mollier chart. 

volume. In the superheated region the long formula for super- 
heated steam would have to be used and by equating this for two 
pressures the degrees of superheat at the second point could 
be found and in this way the curve could be determined. 

The lines of constant S and constant temperature are vertical 
and horizontal lines. 



52 HEAT ENGINEERING 

A diagram such as Fig. 14 is used in practice in which lines of 
constant i, x, v, deg.-sup., s and t are drawn at equal distances 
apart. 

In the Mollier chart, Fig. 15, similar methods are used to com- 
pute various points. The lines shown are constant quality and 
pressure on is coordinates and to this diagram should be added 
lines of constant volume. 

For a definite pressure 



s' + -^ or s ' + -f + 



r at 

and i — q' + xr — Apv' or q f -f r -f- I c p dt — Apv' 



and for different values of x or degrees of superheat at the same 
pressure the values of i and s may be found giving a line of con- 
stant pressure. 

If this is done for several pressures and then the points of the 
same quality are connected the lines of equal quality and equal 
pressure are found. If 



or the equation for superheated steam is used the conditions at 
different pressures for the same volume may be found and, from 
these, lines of constant volumes. Fig. 15 shows this chart. 

Having these charts and formulae, the various lines may be 
discussed. On all lines the important things are the work, heat 
and change of entropy. 

HEAT ON PATHS 

If Fig. 16 represents any line the heat added on it is given by 



JQ = U 2 - C/i + 
I pdv = work 



I pdv 



To find this analytically the equation of the curve must be 
known and if this is not known the area beneath the curve must 
be found by a planimeter and the graphical method used to 
evaluate the integral. If the curve is concave upward, as 



FUNDAMENTAL THERMODYNAMICS 



53 



shown dotted in Fig. 16, it may be assumed to be of the form 
p V n = const., in which case 

P2V2 — P1V1 
1-ri 



/ 



pdv = 



for all values of n except unity; n must be known for this and 
from the pressures and volumes at 1 and 2 it is found as follows: 

PiVi* = p 2 7 a B 



Pi = 
n = 



(W 



log 



£1 



1 F2 

Iog y; 



(121) 



Of course this is the value of n of a curve of this form passing 
between the end points 1 and 2, 
but it may not pass through the 
other points. 

The specific volumes at points 1 
and 2 are now found by dividing the 
volume by M after which these spe- 
cific volumes and pressures are used 
on charts or in tables. 



or 



Fig. 16. — Path onpY plane. 



M(i - Apv) 
M(q' + xp) 

To find this, i or x must be known. 
If from tables or charts i may be 
found, then the first formula is the one to use as it applies to sat- 
urated and superheated conditions. If, however, the table does 
not go low enough x is found by 

V 1 



Xi = 



x 2 



Mv\ 



Of course M must be given in the problem; v" is taken from 
the steam tables. 

Having £/ 2 and Ui their difference is known and from this and 
work the heat is found by addition. 

The change of entropy is determined by finding the entropies 



54 HEAT ENGINEERING 

at 1 and 2 from charts or tables or if these cannot be used they 
must be computed 

, XT 

s = s' + y 

The entropy change is then 

82-81 = M(s 2 - 81) 

This is for any line. For the adiabatic, the heat is zero, the 
entropy change is zero, and the work is equal to the change in 
intrinsic energy. The entropy is constant on this line and this 
gives the equation of the adiabatic 

s\ + x 1 r jr = s f 2 + x 2 7 ~ (122) 

r 2 ( dt 



J 



= S' 2 + jT + I Cp-y 

Tsat. 

Knowing the first point through which the curve must be 
drawn, the unit volume is found by 

V 

V== M 

From tables or charts the quality is found for this specific 
volume. If this cannot be found on account of the limits of 
these tables 

_ v 

X jf 

V 

The entropy of the first point being known, the quality at the 
second point is found by (122) and from this the volume, if needed, 
is 

V = Mxv" or Mv 

If v can be found from tables or charts, it is easier to get it 
from the same entropy column or line. The work is then given 
by 

Work = U 1 - U 2 = M( Ul - u 2 ) 
u = i — Apv or q' + xp 

At times the difference between the u's would lead to such an 
error that it is better to find V 2 for a given p 2 and then compute 
n by the log formula (121) and find work by 

Work = V*V* - P1V1 
1 — n 



FUNDAMENTAL THERMODYNAMICS 55 

The isodynamic is a line of constant U. 

U l = U 2 

Mui = Mu 2 

Now Ui = i — Apv or q' + xp 

.'. ii — A-piVi = i 2 — Ap 2 v 2 or q\ + ZiPi = q' 2 + x 2 p 2 

(123) 

Given the pressure, volume and weight at the first point, the 
quality is found after determining the specific volume by using 
the tables and charts if possible; or by using the formula for 
superheated steam or 

__ v 

x " V 

The value of U\ may then be found and this is equated to u 2 
for the second pressure from which the second quality may be 
computed and then the volume is found by 

V 2 = Mx 2 v" 2 

The heat in this case is equal to the work since there is no 
change of intrinsic energy. The work is found by finding n by 
formula (121) and then 

Work = frr» - pi^i 

1 — n 

The values of s 2 and Si are found and then 

& - Si = M(s 2 - si) 

The isothermal in the saturated region is the same as the con- 
stant pressure line and on this 

Q = M(x 2 - xi)r (124) 

Work = M(x 2 - Xl )f (125) 

S 2 - S! = M(x 2 - xi) ~ (126) 

In the superheated region the curve approaches the rectangular 
hyperbola and it would be difficult to solve problems on this line 
if tables were not available. In this case the formula for super- 
heated steam would have to be used to get the volumes or if 
tables or charts were used the volumes could be found by finding 
the degrees of superheat at various pressures by 

Degree superheat = T — T sat . 



56 



HEAT ENGINEERING 



and this would fix i, v and s. Then 

Work . v^nv\ 

1 — n 
U 2 — Ui = M(i 2 — Ap 2 v 2 — [i\ — ApiVi]) 

Q = U 2 - Ui + work- 
S 2 - #1 = M(s 2 - Sl ) 
On the constant volume line 

Work = 

V 

X " Mv" 
U 2 - Ui = Q = M[q' 2 + x 2P2 - (q\ + x lPl )] 
S 2 - Si = M(s 2 - 8l ) 

FLOW OF FLUIDS 

If a fluid flows through an orifice in which there is a change in 
section there is a drop in pressure and 
as a result there is a change in velocity. 
In Fig. 17 consider the sections 1 and 
2. If the areas are represented by Fi 
and F 2 , the pressures by pi and p 2 , the 
specific volumes by V\ and v 2 , the veloc- 
ities by Wi and w 2 the following consid- 
erations must hold when M pounds of 
substance flow per second. 

Mwi 2 
Kinetic energy at 1 in ft-lbs. = — ^ — 

Internal energy at 1 = Mui 

Work done in pushing substance along = MpiVi 

'Wi 2 




Fig 



17. — Orifice for flow 
of fluid. 



Hence total energy at 1 
Total energy at 2 



= H- 



2g 



+ ui 4- pit>ij 
+ u 2 + p 2 v 2 J 



Now if there is any heat added between these two points, say 
MJq, the energy at 2 must equal that at 1 plus MJq. Of course 
if Jq is taken away the sign would change. 



Hence 



or 



since 



Wi £ 
W 2 



w 2 



+ Ui + P\Vi + Jq = -ft- + u 2 + p 2 v 2 



2g 



2 _ 



Wi' 



2g 



= Jq + Wi + piVi — (w 2 + P2V 2 ) 
= Jq + J(ii 

Ui + P\Vi 



i 2 ) 
Jii 



(127) 



(128) 



FUNDAMENTAL THERMODYNAMICS 



57 



If now F\ is so large that Wi is small and if q = 0, the formula 
becomes 

w 2 ? 



j(< 



in 



or 



2ff 

w 2 = -*j2gj(ii - j 2 ) 



(129) 
(130) 



Since q = this action is adiabatic but in the case that there 
be internal friction, 1 and 2 are not on points on an isoentropic 

line but a line passing to the right of 
this line on the T-S plane. In other 
words, i 2 is greater than it would have 
been if there had been no friction. 
The amount of this increase in i 2 is 
usually found by assuming that 2 has 
the same entropy as 1 and then in- 




Fig. 



18. 



-TS d 

of flu 




agram for flow 
ds. 



Fig. 19. — pV diagram for flow of 
fluids. 



stead of using %\ — i 2 for these points, only a fraction of this is 
used. In Fig. 18 abcde is equal to i\, since 

abcf = q' 

cdef = xr 
abcde = q' + xr = i — Ayv' = i 
abg2e = i 2 
i\ — i 2 = cd2g 

This assumes no friction, hence d and 2 have the same entropy. 
By experiment the heat used in friction is found to be y(ii — i 2 ), 
hence this must be subtracted to get the amount of heat left to 
give the gain in kinetic energy. Hence 

w 2 = ^2gJ(i 1 - i 2 ){\ - y) (131) 

i\ and i 2 are for points on the same entropy line. In Fig. 18 the 
area cd2g is cut down by the amount hd2k which is equal to 



58 HEAT ENGINEERING 

y(ii — ^2). This heat stays in the substance and hence at exit i 
is not the i of 2 but that of 2' which is fixed by making 

2Ve2 = M2fcor22' = ^P 

1 2 

If this value of iy could have been determined the original 
formula (130) would have been used. It is because i v cannot 
be found that this method of using a portion of the amount for 
isoentropic expansion is employed. 

It must be remembered that the shaded area less the friction 
loss equals the gain of kinetic energy. 

— 2~ — = J(n - *a)(i - y) 

The quantity i\ — i% is the same as the area behind the adia- 
batic on the pv plane as shown in Fig. 19. 

Area al2c = aleb + el2d — c2db 

= P1V1 + (u'i — u 2 ) — V2V2 
= J(ii — i 2 ) 

On account of friction this is reduced as pointed out above. 

THROTTLING ACTION 

Suppose now that there is so much friction that there is no gain 
in kinetic energy although there is a drop in pressure; this is 
called throttling action. In this case 

w 2 2 Wi 2 

~2g-~2g ={) = Jfe " ^ 

or i x = %2 (132) 

This means that in throttling action the heat content is con- 
stant. The point 2' of Fig. 18 then is found to be on a curve of 
constant heat content. The horizontal lines of the Mollier 
chart or the constant i curves of the T-S diagrams are throttling 
curves on these diagrams. 

Since in perfect gases 

Ji =k~i V1V1 = ^n MBT ^ ( 133 ) 

the throttling curves for such are curves of constant temperature. 



FUNDAMENTAL THERMODYNAMICS 59 

VELOCITY OF VARIOUS SUBSTANCES 

For short tubes and orifices the friction is negligible and the 
action may be considered isoentropic as well as adiabatic. 



w 2 = ^2gJ(t! — i 2 ) 
For liquids, since the temperature change is slight, 

i = Apv + const. 



w 2 = V2#[piFi - p 2 V 2 ] = V2flf[pi - p 2 ]V 
Since Vi = V 2 

7 = 1 

m 

[pi ~ V2]— = h 
m 

w 2 = yl2gh (134) 

This is the usual formula from hydraulics for the velocity of a 
liquid. 

k 
For gases Ji x = , _ . piVx 



w 2 =^ Zg^zTi (i >lVi ~'P 2V2 ) = 



V 2 ^^ 1 -©^"]^ 



jfc-i 

Since piVi k = p 2 v 2 

This is the formula for velocity and is a maximum when p 2 = 0. 

DISCHARGE FROM ORIFICES 

Now Mv = volume per second = Fw 



This is zero when p 2 = Pi and when p 2 = 0. Of course the 
latter is unthinkable because there will always be some weight 



60 HEAT ENGINEERING 

when p 2 is less than pi. The explanation is that when the pres- 
sure p 2 falls the pressure which exists in the plane of the orifice 
where the area is F can never become less than that to give maxi- 
mum discharge. To find the condition for a maximum value 
of M, the variable part of the expression is differentiated. For 
a fixed pi and v\ 



©*■-© 



k+i 

h 







/ 2 \ — 

p'-^U + ir" 1 


When h = 


= 1.4 for 


gases this becomes 
p 2 = 0.5283pi 


while for steam, k = 


1.135, and 






p 2 = 0.574pi 



is differentiated with regard to the variable p 2 and by equating 
this to zero, the condition for a maximum is 

(137) 



(138) 



(139) 

When p 2 becomes less than the critical value given above for a 
short mouthpiece, the pressure at this point of area F remains 
at the critical value. Hence the discharge is constant for all 
values of p 2 below the critical value. For air this has been proven 
to be true experimentally by Fliegner who proposes 

M = 0.53F-^ (140) 

when p 2 < 0.53pi 

and 



M = 1.06Fy l P * (pi T - pi) (141) 

when p 2 > 0.53pi 

Equation (140) is really a reduction of (136) by substituting 
(138) for p 2 and reducing. 

For steam similar experimental results have been found by 
Napier. Rankine reduced these results to the form 



Fp 
70 

when p 2 < 0.574pi or 0.6pi 



M = ^ (142) 



FUNDAMENTAL THERMODYNAMICS 


61 


M= F /*J 3( V P2) 
42 \ 2p 2 


(143) 


when p 2 > 0.6pi 




The general equation (131) may be used for the 
shown above and then M found by 


velocity as 


M = F T 





after v"i is found for the condition at outflow. 



CHAPTER II 
HEAT ENGINES AND EFFICIENCIES 

A heat engine is any machine in which heat is used to furnish 
the energy for the production of mechanical work. As examples, 
the steam engine, the steam turbine, the gas engine and similar 
machines may be mentioned. 

To make heat available in one of these engines it must be 
applied to some substance which undergoes changes. These 
changes form a cycle and the changes which affect the properties 
of the substance may be studied on planes of projection by show- 
ing the successive values that certain properties take. The 
paths of the change which the substance undergoes are known as 
a cycle. 

As pointed out in the chapter on Fundamental Thermo- 
dynamics Qi heat units are supplied and Q 2 heat units are re- 
jected, giving Qi — Q 2 units of work. 

All of these machines work between some range of temperature, 
Ti to T 2 , and consequently the availability of the heat is 

Ti W 

This is the only part of the heat which could be turned into 
work and therefore represents the highest possible efficiency. 
It represents the efficiency of the Carnot cycle for this tempera- 
ture range and therefore it is sometimes spoken of as the Carnot 
efficiency of a given cycle. Of course any cycle would have to 
be a Carnot cycle to have this efficiency, but all that the term 
means with reference to a particular cycle is the maximum value 
that might be possible for this cycle working between Ti and T 2 . 
Calling this 771 

vi - — f ~ ~T ^ ' 

If Qi is the theoretical amount of heat added and Q 2 the theo- 
retical amount rejected, which cannot be used, 

Q 



= Vz \ 6 ) 



62 



HEAT ENGINES AND EFFICIENCIES 63 

This is the theoretical efficiency of the cycle. 

The ratio of 773 and 771 shows how close the efficiency of the 
theoretical cycle approaches the maximum possible efficiency of 
the cycle. This is called the type efficiency, 772 

„ = * (4) 

If this is nearly unity it indicates that the theoretical cycle 
is almost equal to that of Carnot and hence in theory the cycle is 
good while a low value of 772 shows that the cycle is a poor one 
theoretically. For example, the steam engine cycle has a value of 
772 above 0.90 while in the case of the gas engine 772 may be less 
than 0.50. These indicate that an improvement may be ex- 
pected in the type of cycle used in a gas engine, although for the 
steam engine there is little hope of bettering the cycle. 

If for an amount of indicated work AW a the amount of heat 
Q a is actually required, the actual thermal efficiency 775 is given 
by 

AW - m 

If the heat supplied per unit of substance used is q a , be it coal, 
steam, gas or air, and if M is the amount of substance per 
indicated horse-power hour, this expression becomes 

AW a 77^ X 33000X60 
Q a Mq a Mq a V5 

2546 B.t.u. = 1 h.p.-hr. 
42.43 B.t.u. = 1 h.p.-min. 

The ratio of 775 to 773 is called the practical efficiency, 774, and 
shows how near the actual efficiency approaches the efficiency 
demanded by theory and a low value of this means that there 
have been errors in actually applying the cycle. To make this 
term larger, jackets, superheated steam, and reheaters have been 
applied to steam engines. 

* - ir ^ 

If the mechanical efficiency 776 is the ratio of the output to 
that developed within the machine or shown by the indicator 
card, this efficiency is found by 



64 



HEAT ENGINEERING 



VG = 



output 



indicated work 



AWo 
AW a 



(7) 



The overall efficiency is then the product of certain of these 
various efficiencies. 

output 



V = 



(8) 



heat supplied 

The overall efficiency being the product of these, the efficiency 
may be increased by increasing any of them. In cases such as the 
gas engine 771 is so great that although 772 and 77 6 are small the 
product is greater than that of the steam engine. These various 
efficiencies will be investigated for different machines. 




Fig. 20. — pV and TS diagrams of cycles. 

The quantity A W a may be found from the area of the cycle on 
the pv plane or from the area on the T-S plane in Fig. 20, if the 
lines are reversible lines. Of course on the pv plane 123456 
represents positive work while 6571 represents negative work 
and hence 723457 represents the network. 

On the T-S plane, 123456 represents Qi if there is no friction, 
and 6571 represents Q 2 under these conditions, hence the area 
of the cycle represents Qi ~- Q2 or the work done. If, however, 
there is friction 

12345 = Qi + H 1 
and 6571 = (- Q 2 + H 2 ) 

since the heat removed to the outside from 5 to 7 is more 
than that shown by the area on account of the heat developed by 
friction. 

23457 = Q 1 - Q 2 + (#1 + #2) = AW + #1 + H 2 



HEAT ENGINES AND EFFICIENCIES 



65 



or the area is greater than the work of the cycle. In any case 
an irreversible line makes the work less than the area of the cycle 
on the T-S plane. 

Looking at the Carnot cycle of Fig. 21, or the cycle of Fig. 20, 
it is seen that the efficiency is 



2345 
123456 



Qi-Q* 
Qx 



If in the Carnot cycle heat is not added on 34 but on some other 
line 3'4 or 34', the heat added and work developed are decreased 
by the area of the small triangle. 

2345 - 33'4 



Hence 



Eff. = 



123456 - 33'4 



Subtracting the same thing from numerator and denominator 
of a fraction less than unity decreases the value of the fraction. 
This may be seen by remembering that the effect of the removal 
of the triangular area is greater on the smaller area. Hence 
the efficiency is decreased. 















3 

3' 


— "~""~"~ — -— ~. 


4 
4' 

5" 






2 




5 










T 














> 


6 


S 



Fig. 21. — Conditions for maximum efficiency. 

If the heat were removed on a line 5 "2 of Fig. 21 this would cut 
down the work but would not effect the heat, hence the efficiency 
would be diminished in this case. 

From the above it may be said that for a given range of tem- 
perature heat must be added or taken away at constant tempera- 
ture if the maximum efficiency is to be obtained and since this 
efficiency is 

r, 

the values of T 2 and T\ (the limiting temperatures of the range) 
are to be separated as much as possible. 

An important point must be borne in mind. Although the 



66 



HEAT ENGINEERING 



statement above is absolutely true, it does not follow that a 
cycle in which heat is added with a varying temperature finally 
reaching a high value may not be more efficient than one in which 
heat is added at a constant temperature of lower value. Such 
cycles may be made necessary by the nature of the medium used. 
This is shown in Fig. 22. 1234 is less efficient than 5678. What 
is meant is that given the highest and lowest possible tempera- 
tures of a cycle, the greatest efficiency would be obtained if all 
of the heat added were added at the highest temperature and all 
of the heat removed were abstracted at the lowest temperature. 
Thus in Fig. 23 the efficiency of the steam engine with saturated 

steam is — rr while that with superheated steam is — ttt- • In 
eabf eabb g 

the figure — rr is practically equal to — r — and by adding the 







7 


2 




3 








e 






i 




4 


5 




8 



i'l 1 

i / \ / 


h 


! l a 




Id 

<\ 

1 
i 

! 
1 

i 


c \ C 
\l 

l\ 

1 
1 
1 

! s 



Fig. 22.— Cycle with 
varying temperature on 
heat line. 



e f g 

Fig. 23. — Steam engine 
cycles with saturated and 
superheated steam. 



triangle bb'h to numerator and denominator the expression for 
the superheated cycle is obtained. This addition having a 
greater effect on the numerator increases the efficiency. Hence 
the use of superheated steam, although the heat is not added at 
constant temperature, does increase the efficiency a slight amount. 
For other reasons than those mentioned here the use of super- 
heated steam increases the efficiency of the engine. The effi- 
ciency would be increased by a greater amount if the heat could 
have been added on the dotted line &'b f . In this case the effi- 
ciency would have been 

d'b'c'd 

ed'b'g 



HEAT ENGINES AND EFFICIENCIES 67 

In some machines as the steam turbine the indicated work 
cannot be obtained and in such a case the product 775 X rje only 
can be determined. 

METHOD OF REPORTING PERFORMANCE OF ENGINES 

r 

Although the efficiency of an engine tells an exact story, for 
commercial reasons it is quite common to report the performance 
of a heat engine in pounds or cubic feet of substance required 
per indicated or brake horse-power hour. Thus pounds of steam 
per horse-power hour for an engine or turbine, cubic feet of 
standard gas per horse-power hour for a gas engine, coal per horse- 
power hour for a gas or steam engine have all been found in re- 
ports of tests. This data on the horse-power hour or kilowatt- 
hour basis is valuable for commercial reasons but because differ- 
ent gases and coals have different heating values, and steam at 
different pressures and qualities contains different amounts of 
heat, these statements are not definite until other data are known. 

When coal is used in a boiler or producer the coal per unit of 
output depends on the efficiency of the boiler or producer as well 
as upon the engine. To separate the losses and find out just 
where losses occur and just what they are, it is always better to 
give the efficiency of the heat engine as a percentage using the 
heat supplied as the base. These other methods are valuable 
for commercial purposes and the quantities should be reported. 

A method used in reporting the performance of pumps is by 
duty. This is the amount of useful work in foot-pounds per (a) 
100 lbs. of coal, (b) 1000 lbs. of dry steam, or (c) per 1,000,000 
B.t.u. The first two methods are not definite although the third 
method is exact. The duty by the third method, divided by 
778,000,000, will give the overall efficiency. 

Results of a number of tests will now be given: 

HIGH-SPEED NON- CONDENSING ENGINE 



I.h.p 

B.h.p. 


. 130 
. 120 

. 115.3 lbs. per 
sq. in. gauge 


347.4-213 ,„„ 

7?1 -347.4+460 _16 - 7% 

1190-1078.5+~|(29.8 -15)12.58 


Steam pressure . . 


//3 ~ 1190-181.3 

= 14.5% 
?72=m=86.7% 



68 HEAT ENGINEERING 

Barometer 14.7 lbs. per _ ___2546 

sq. in. r?5 ~ 30.5[1190- 181.3] ~*- 6 '° 

Quality of steam . . 1 . 00 _ 8.3 _ 

Pressure at end of expan- 120 

1c1 u • 776=^^ = 92.5% 

sion. . . 15 lbs. per sq. m. gauge 16[J 

Back pressure. . . . 0.3 lb. per sq. 

in. gauge 

Steam per i. h.p. hr. 30.5 lbs. 

42.42 
B.t.u. per l.h.p.-min. =7y^o = 510 

HIGH-SPEED NON-CONDENSING COMPOUND ENGINE 

I.h.p 130 7? 1 = 16.7% 

B.h.p 115 773 = 14.5% 

Steam pressure 115.3 lbs. per sq. 772 = 86.7% 

in. gauge 

Barometer 14.7 lbs. per sq. in. 2546 11707 

?76 ~21.5x[1190-181.3]~ 11 - //o 

Back pressure 0.3 lbs. per sq. in. H-? Q1 

gauge j/t.o 

Pressure at end of ex- _ H5 

pansion. . . 15 lbs. per sq. in. gauge 130 

Quality of steam .... 1.00 

Steam per i. h.p.-hr. . 21.5 lbs. 

42.42 
B.t.u. per l.h.p.-min. = .. 7 =362 

LOCOMOBILE ENGINE 

I.h.p 191 391+282-139 

971 "391+ 282+ 460 ~' /o 

Kw. generator 121.5 1352- 1002 

m "1352- 107 -^• 1 % 

Steam pressure 208 lbs. per sq. in. (Complete expansion.) 

gauge 
Feed Temperature. . . 132° F. 28^1 

Barometer 14.7 lbs. per sq. in 

9.9[1352- 107] 

Vacuum 11.9 lbs. persq. in. 20.6 

r?4 = 28~T 

Degrees superheat . . 282° F. v 121.5 _ 

776X77 7 - 191x0746 -S5.o/ 

Steam per i.h.p,hr.. 9.9 lbs. Eff . of boi1er = 8.28[1352-100] = 73 6% 

Steam per lb. coal . . . 8.28 lbs. 

Heat of coal 14,099 B.t.u. per lb. 

42.42 
B.t.u. per l.h.p.-min. = a 206 = 2 ^ 



, 5 = ,^W20.6 % 



HEAT ENGINES AND EFFICIENCIES 69 



PUMPING ENGINE 

I.h.p 861.34 384.2-108 

771 "384.2+ 460 ~ 6ZJ% 

Deh h -P 839 - 8 1189-926+^(4-1.2)67.8 



VZ = 



1189-76.0 

= 26.7% 



Steam pressure 190.8 lbs. per sq. ^ 26.7 

in. gauge 32.7 

Barometer 14.8 lbs. per sq. in. 2546 991w 

r75_ 10.37[1189-763"^- 1/o 

Back pressure 1.2 lbs. per sq. in. 22A 

abs . , 4 = 267 7 = 82.8% 

Pressure at end of ^ 839.8 

expansion 4 lbs. persq. in. abs. 861.3 

Quality of steam .... 0.99 
Steam per i.h.p.-hr . . 10.37 

42.42 
B.t.u. per l. h.p.-min. = n ooi =192 

Duty per million B.t.u. =778,000,000X0.221 X0.974 
= 167,000,000 ft.-lbs. 

STEAM TURBINE 

Kw 6257 549-76 

,7l = 559+460 = 45 ' 6% 

Steam pressure 203.7 lbs. persq. in. 1290—879 

abs. " = 1290-44 = 33% 

Superheat 165.5° F. 33 

" 2 = 4K6 = 72 ' 5% 

Barometer 29.92" 2546 

175 ~ 11.95 x[1290-44] X0.746 

= 22.9% 

Back pressure 0.44 lbs. per sq. in. 22.9 

, ^ 974 = -^- = 69.5% 

abs. && 

Steam per kw.-hr .. . 11.95 1b. 

42 42 
B.t.u. per kw.-min. = .746x 0.229 = 249 

B.t.u. per elec. h.p.-min. =249X0.746 = 186 

PRODUCER GAS ENGINE 

I.h.p 579 2546 

775 ~ 0.738 X 0.805 X 14320 
= 30% 

B.h.p 483 483 00 „ 

776 = 579 = 83 ' 5% 
Coal per i.h.p.-hr... 0.805 

Heating value of coal 14,320 B.t.u. per lb. 



70 HEAT ENGINEERING 

Kind of coal Bituminous 

Efficiency of producer. 73.8% 

42.42 
B.t.u. per l.h.p.-min. = "ttoTT = 141.4 

141.4 
B.t.u. per b.h.p. -min. = ttook = 169 

BLAST-FURNACE GAS ENGINE 

I.h.p 775 2546 

7,5 = 110.5 X905 = 25 ' 5% 

B.h.p 565 565 

ve = ^ = 73% 

Cu. ft. of gas per i.h.p.-hr. 90.5 

Heat value of gas 110.5 B.t.u. per cu. ft. 

42.42 
B.t.u. per i.h.p. -min. = ~ ~_- = 166 

1 f\f\ 
B.t.u. per b.h.p. -min. = ^ry^ = 227 

DIESEL OIL ENGINE 

I- h -P 523 2546 

Vb ~ 0.37 X 19270 ~ d& '* % 

B.h.p 450 _450_ 86 c7 

776 " 523 " 8b% 
Oil per i.h.p.-hr. . 0.37 lb. 
Heat value of oil. . 19,270 B.t.u. per lb. 

42.42 
B.t.u. per i.h.p.-min. = tToko — 118 

118 
B.t.u. per b.h.p.-min. = — — = 143 
0.83 

RESULTS OF VARIOUS TESTS 

Engine tested 

110-h.p. Nurnberg gas engine on coke 

110-h.p. Nurnberg gas engine on anthracite coal. . 

210-h.p. Guldner engine on illuminating gas 

600-h.p. Ehrhardt engine on coke-oven gas 

11,000-kw. Westinghouse turbine 

11,000-kw. Curtis turbine 

1,500-h.p. locomotive 

10,000-h.p. marine engine 

2,200-h.p. Corliss engine 

1,000-h.p. air-compressor engine 



B.t.u, 

per i.h.p- 

min. 


B.t.u. 

per b.h.p- 

min. 


110 


138 


120 


150 


100 




113 


136 




213 




201 


350 




246 




226 




169 





TOPICS 

Topic 1. — What is a heat engine? What is a cycle? What is the general 
expression for the efficiency of any cycle? What is the expression for the 



HEAT ENGINES AND EFFICIENCIES 71 

efficiency of the Carnot cycle? What does this efficiency represent? Give 
the meaning of the terms: Carnot efficiency, type efficiency, theoretical 
efficiency, practical efficiency, actual thermal efficiency, mechanical effi- 
ciency and overall efficiency. 

Topic 2. — Give the meaning of the symbols: 771, 772, 173, va, vb, vg, v- Give 
the relations between these and the formulae by which each is found. Tell 
the manner of determining the quantities entering into these formulae. 

Topic 3. — Give the conditions for maximum efficiency in a heat engine and 
show that although these conditions may not be fulfilled high efficiencies 
may be obtained. Are these high efficiencies as high as they would be were 
the conditions for maximum efficiency fulfilled? How are results of tests 
reported? 

PROBLEMS 

Problem 1. — An engine using 35 lbs. of steam per h.p.-hr. at 125 lbs. gauge 
pressure, x = 0.98, and with a back pressure of 2.5 lbs. gauge, has its con- 
sumption reduced to 30 lbs. when supplied with steam under the same 
pressure but superheated 245° F. The back pressure does not change. 
The barometer is 29.8 in. What is the per cent, saving, if any? 

Problem 2. — A Corliss engine gives the following test results: i.h.p. = 
135.6; b.h.p. = 126.0; steam per hour, 3406 lbs.; pressure at throttle valve 
by gauge, 135.6 lbs. per square inch; barometer, 14.68 lbs. per square inch; 
pressure at end of expansion by gauge, 20.5 lbs. per square inch; back pres- 
sure by gauge, 1.5 lbs. per square inch. Find the various efficiencies, 77, 776, 
775 and 771. 

Problem 3. — A Corliss condensing engine using steam at 174 lbs. gauge 
pressure with x = 0.995 and a vacuum of 27 in. with a barometer of 29.8 in. 
consumes 17.5 lbs. of steam per kw.-hr. output from generator. Find 
overall efficiency. By installing a low-pressure turbine and by reducing 
the vacuum to 28.5 in. and superheating the steam to 200° F. of superheat 
this consumption is made 13.8 lbs. of steam per kw.-hr. and the output of 
the plant has been increased 75 per cent. Is the change of value? Why? 

Problem 4. — A pumping engine gives a delivered duty of 175,000,000 ft.- 
lbs. per 1,000,000 B.t.u. when supplied with steam at 165 lb. gauge pressure 
with x = 1 and a temperature of hot well of 105° F. How many pounds of 
steam are used per delivered horse-power hour? If the mechanical efficiency 
is 95 per cent., what is the steam consumption per i.h.p. -hour? 

Problem 5. — A gas engine has an overall efficiency of 24 per cent. The 
heat lost in the jacket is 25 per cent, of that supplied by the gas. How 
much heat is added to the jacket water per hour if the delivered power is 
565 kw.? 

Problem 6.' — The results of the test of a producer gas engine plant are as 
follows: Coal burned in 120 hr., 16,000 lbs.; heating value of coal, 14,450 
B.t.u.; gas produced during test, 1,500,000 cu. ft.; heating value of gas, 120 
B.t.u. per cubic foot; b.h.p., 120; i.h.p., 150; cooling water, 1,000,000 lbs.; 
temperature of inlet water 65° F., outlet water 110° F. Find the efficiency 
of the producer. Find the overall efficiency, 77. Find the indicated 
efficiency, 775. Find the heat removed by the jacket water. Find the heat 
per b. h.p.-hr., per b.h.p. -min., per i.h.p. -hour and per i.h.p.-min. 



CHAPTER III 
HEAT TRANSMISSION 

The phenomenon of the transmission of heat through partitions 
is very important as it enters into the determination of the areas 
of radiators for heating systems; of the surface required in con- 
densers, boilers, evaporators, intercoolers, and for many other 
engineering structures. Its consideration is important in finding 
the heat required for warming a building or the amount of re- 
frigeration to keep a certain cold storage warehouse at a low 
temperature. The transmission of heat is complicated and al- 
though much experimentation has been done the exact laws are 
not completely determined. The following discussion is based 
on the works of various authors. The results of the authors 
have been arranged so as to give the student a working knowledge 
of this important part of applied thermodynamics. 

To transmit heat from one body to another or from one part 
of a body to another it is necessary to have a difference of tem- 
perature. Having this, the heat may be transmitted by one 
or more of the three methods : radiation, convection and conduc- 
tion. Radiation is the method of transmitting heat by vibrations 
of the ether. The hot body starts this vibration which is trans- 
mitted in all directions through the ether and until another body 
receives the vibration the energy is not sensible. As soon, how- 
ever, as a body opaque to these vibrations is placed in the path 
it becomes heated. In convection heat is applied to some mov- 
able body and then this energy is conveyed by the actual move- 
ment of the body with its heat. In this method heat is carried 
by moving particles of matter. Conduction is the method by 
which heat travels from the hot portions of a body to another 
part which is colder. In this method it may be that the more 
violent vibrations of the particles of the hot portion of the body 
are gradually transmitted to those in less violent vibration 
(colder) or according to the later views of the constitution of 
matter it may be that more electrons are thrown off at higher 

72 



HEAT TRANSMISSION 73 

temperature vibrations and these gradually affect the vibrations 
of the atoms at the more remote but colder portions of the solid. 
The laws of transmission by radiation are well known experi- 
mentally and theoretically. The Stefan-Boltzmann Law for the 
radiation from a black body is 

q = CF(TS - 7Y) (1) 

Q = amount of heat per hour in B.t.u. 

C = a constant = 1.6 X 10~ 9 . 

F = area radiating heat, in square feet. 

Ti = absolute temperature in degrees F. of hot body. 

T 2 = absolute temperature in degrees F. of cold body 
which must surround the hot body or which 
must include all rays issuing from the hot body. 

It is to be remembered that the formula applies to black bodies 
and then only when the cold surface includes all rays from the 
hot body. If the surface subtends a solid angle co when it should 
have subtended a solid angle of 2x or a hemisphere, the quantity 

Q is found by multiplying (1) by — • 

Zir 

The law of radiant energy has been under discussion from the 
time of Newton in 1690, who proposed a law proportional to the 
first power of the temperature, until 1879 when Stefan proposed 
that the law be of the form given in (1) basing his assumption 
on some experiments of Dulong and Petit, Tyndall and others. 
After some years (1884) Boltzmann proved that this form was 
correct theoretically. A black body has to be assumed since a 
colored body would reflect a certain amount of energy and thus 
the amount emanating would include this in addition to the 
amount radiated. For bodies which are not truly black (absorb 
all radiation falling on them) the law is approximately true. 
The nature of the surface has also an effect and this should be 
taken into account. Polished bodies will not radiate as much 
as rough surfaces. 

The above law may for convenience be put in the form 

^ 160 °[(lSo) 4 -(lSo) 4 ] <» 

The above refers to a black body of perfect radiating power. 
The relative values of different surfaces by Lucke are given in the 
following table: 



74 HEAT ENGINEERING 

Absorbing 
power 

Porous carbon 1 . 00 

Glass 0.90 

Polished cast iron . 25 

Polished wrought iron . 23 

Polished steel . 19 

Polished brass . 07 

Hammered copper 0.07 

Polished silver 0.03 



The transmission of heat by pure conduction is a simple 
matter. The law for this is similar to that of the transmission 
of electric current. The flow of heat is proportional to the differ- 
ence of temperature and the area of the substance and is inversely- 
proportional to the length of path or thickness. Thus 

CF 

Q = -j-Xh-t 2 ) (3) 



i Q = B.t.u. per hqur. 
F = area in square feet. 
I = thickness in feet. 
t\ = temperature at one side in degrees F. 
t 2 = temperature at one side in degrees F. 
Fig. 24— C = coefficient of conduction in B.t.u. per hour 
siorf surface" P er S( l uare ^ 00 ^ P er degree F. for 1 ft. thickness. 




Values of C 
C = C [l + a(t - 32)] 





t = temperature 


in degrees F. 






Substance 


Co 


a 


Substance 


Co 


a 


Air 


0.03 to 0.012 
53.5 
0.46 
0.8 
0.006 
40.5 
0.46 
239.0 
0.17 


0.0011 
0.0011 

-0.00012 
0.00003 


Cork powdered 
Glass 


0.03 
0.54 

48.5 
1.35 
0.46 
0.87 

26.7 
0.292 
0.10 




Brass 




Brick 


Iron 


-0.00012 


Carbon 

Carbon dioxide 

Cast iron 

Concrete 

Copper 

Cork board. . . 


Limestone 

Masonry 

Sandstone 

Steel soft 

Water 

Wood 





HEAT TRANSMISSION 75 

The value of C for gases has been shown by Maxwell 1 to be 
given by 

C = Kr)C v 

K = constant between 0.5 and 2.5. 
7} = coefficient of viscosity = the force between two 
planes separated by distance unity when one plane 
is moving with unit velocity relative to other. 

c v = specific heat at constant volume. 

He also showed that C varies as the % power of the absolute 
temperature. 

Nusselt examined insulating materials and found that the 
conductivity of these substances increased as the absolute 
temperature. 

The amount of heat carried by convection depends on the 
amount of substance involved, its specific heat and its tempera- 
ture. This is simple to find but the amount of heat which can 
be abstracted from these particles by a given surface under given 
conditions is a complex matter and it is this method which will 
be considered throughout this chapter. This transfer of heat 
through partitions is the important one in most applications of 
heat. 

When heat is transmitted through a surface as in a boiler or as 
in an indirect heating coil there must be films of substance on each 
side of the wall which prevent the passage of heat. Suppose, 
for instance, in a boiler where the temperature of the hot gases 
on one side is 1500° F. and the temperature of the water on the 
other is 325° F., the thickness of the steel tube is ^ in. and that 
there are 4 lbs. of water evaporated per hour per square foot 
under these conditions. 

The heat transmitted per square foot per hour 

= 4 X r 325 

= 4 X 889.8 = 3559.2. 

For the steel wall, by the law of conduction, there results : 

26 7 
3559.2 = ly ' X (*i - U) 

* - «»- ifflb - 2 - 78 ° F - 

1 Phil. Mag., 1860. 



76 HEAT ENGINEERING 

Now the difference between the water and the gas is 1175° F. 
and of this there is only a drop of 2.78° F. in the steel wall. A 
thin scale of J^oo m - on the water side and }/{ $ in. of soot on the 
gas side would account for considerable drop. 

For the scale, C will be taken as 1. 

_ 35 59.2 _ 2( . rv 
U ~ 1 2 - 1200 X 1 - 2 - 97 ^ 

For the soot, C will be taken as 0.1. 

t' 1 ~ tl = 16X12X0.1 = 185 ° F * 

This accounts for 188° of the 1175°, showing that there must be 
still further resistance at the surface. This of course must be 
due to films of water and of gas. It is seen from the tables of 
values of C that water has about thirty times the value of C for 
air and gas, and hence there will probably be Ml of 987° drop in 
the water film and 3 %\ of 987° drop in the gas film if these are the 
same thickness. The gas is probably less thick and if the water 
film is taken as three times the thickness of the gas film the drops 
in each will be approximately 90° in the water and 897° in the gas 
film. 

The thicknesses to give these results are found below assuming 
C for water 0.292 and 0.009 for the gases. This is the mean of 
0.006 and 0.012. 

292 V 90 

Later = 3559 3 X 12 = 0.089 111. 

U- ^^f- 7 X12 = 0.0272 in. 

Thus it is seen that a film of small thickness on either side 

accounts for a great drop which must exist to explain the low 

conductive powers of the heating surface. If it were not for 

the films and scale the heat transmitted by the steel tube would 

be 

26 7 
Q = -r-r-(1500 - 325) = 1,505,000 B.t.u. per hour. 

74,8 

This would mean an evaporation of 1690 lbs. of water per 
square foot of heating surface. Fig. 25 shows the temperature 
gradient across these surfaces. It appears from the figure and 
the calculation above that anything which would tend to decrease 



HEAT TRANSMISSION 



77 



the thickness of the films of water or gas would cut down the 
drop of temperature in these, putting a greater drop in the wall 
and so causing an increase in the heat transmitted. One method 
of doing this is to increase the velocity of the substance bringing 
heat up to the surface or taking heat from the surface. This has 
been tried and found true. The increase of velocity of the gas 
or water wipes some of the film away decreasing its thickness. 
This simple explanation is evident and shows why one would 
expect an increase in the heat transmitted, if either the velocity 
of the gas or the velocity of the water 
were increased across the surface 
transmitting the heat. 

In October, 1909, Prof. W. E. Dalby 
presented to the Institute of Mechan- 
ical Engineers of Great Britain a paper 
on Heat Transmission, "the purpose 
of which was to place before the mem- 
bers of the Institution a general view 
of the work which has been done re- 
lating to the transmission of heat 
across boiler- heating surfaces." He 
has given this view and with it a list 
of 406 references to various papers 
dealing with this subject. They are 
listed chronologically, by Authors and 
by Subjects, and the student is re- 
ferred to this paper for most of the lit- 
erature on this important subject. 

The fact that the velocity of the 
gases affected the rate of transmission 
was experimentally known as early as 1848 but there does not 
seem to be a statement of this until 1874 when Prof. Osborne 
Reynolds read a paper before the Literary and Philosophical 
Society of Manchester (Vol. xiv, 1874, p. 9) 1 "On the Extent 
and Action of the Heating Surface for Steam Boilers." In this 
very short paper of a few pages, Reynolds points out the im- 
portance of the subject. He states that the heat carried off 
from the gas is proportional to the internal diffusion of the fluid 
at or near the surface, that is, on the rate at which the particles 




Fig. 



25. — Temperature 
gradient. 



See also The Steam Engine, by John Perry, p. 594. 



78 HEAT ENGINEERING 

which give up their heat diffuse back into the hot gas. This ac- 
cording to him depends on two things: 

1. The natural internal friction. 

2. The eddies caused by visible motion which mix up the 
fluid and continually bring fresh particles into contact 
with the surface. 

The first of these is independent of the velocity, and the second 
term is dependent on the velocity and density of the fluid, 
the heat transmitted at any point being 

Q = At + Bmwt (4) 

where Q = heat per square foot per hour 
A, B = constants 

t = difference of temperature 
m = density of substance, pounds per cubic foot 
w = velocity along surface in feet per second. 

He then discusses the quantities A and B and the way in which 
the heating surface may be of more value by an increase in 
velocity. He finally closes his paper with the hope of giving a 
further communication. He does not give the values of A and B. 

In Reynolds' paper he gave no theoretical discussion nor deri- 
vation of his formula. 

In 1898, Professor John Perry gave the following theoretical 
discussion to prove that transmission varied as the velocity : 

In a thin film the number of molecules, n, entering the film per 
square foot of area is equal to the number leaving and the friction 
force developed by the axial momentum given up by the n mole- 
cules as they are arrested from velocity w will be given by 

P(per square foot) <x nw (5) 

Now since the kinetic energy is proportional to the tem- 
perature the change in the energy as the surface abstracts heat 
will be 

Q oc n (t - 6) (6) 

t = original temperature of gas 
d = temperature at surface of pipe. 

Eliminate n from (5) and (6) 

Q « £('-«) (7) 



HEAT TRANSMISSION 79 

It is known that the force of friction expressed in feet head 
when a fluid flows through a pipe is proportional to the square 
of the velocity w 2 . To change this to pounds per square foot it 
must be multiplied by the weight of 1 cu. ft., m, or 

P oc mw 2 (8) 

Hence (7) reduces to 

Q = K'mwii - 6) (9) 

where K' = constant of equality. 

If now, Perry points out, there is a film of gas at the side of 
the surface of thickness b and conductivity K, there will really 
be a drop from f to 9 in the film and only. a drop from t to t' 
in the gas so that 

Q = K'mwit - t') = ~ (T - 6) 



t' 



rr 

K'mwt + -T- 6 
-j- + K'mw 



or Q = K'mw 



K'mwt + ^ 

z 



j- + K'mw 



^t^ 1 (10) 

bK mw 

1 + ^W 



K' 

If rwi be called K" the formula will be the same as 

oK mw 
1 +~^— 

(9) with a different coefficient. If b decreases as w increases, 
which is probable, the product bw will be nearly constant and 
since K is large and m is small for most gases the denominator 
of the fraction will change little. Hence formula (9) is practically 
correct. 

This equation of Perry and that of Reynolds may be changed 
by remembering that 

M = mFw 

where M = total weight per second in pounds 
F = area of passage in square feet 
w = velocity in feet per second 
m = weight per cubic foot. 

TT M 

Hence -^ = mw 



80 HEAT ENGINEERING 

M 
and Q = At + B -^t 

Q = K"f(t - 6) (10') 

In Perry's formula the fact that P varied with m and w 2 
might have been made to include the fact that P also varies 
inversely with the mean hydraulic depth, di. Thus: 

„ mw 2 

The hydraulic depth or hydraulic radius is equal to the area 
of the cross section of a passage carrying a fluid divided by the 
perimeter of the cross section. 

Hence Q = K'— (t - 6) (11) 

di 

This formula states that the smaller the tube the greater the 
heat transmission. 

Dalby shows that in 1888 Ser and in 1897 Mollier gave for- 
mulae in which the heat transmitted depended on the square 
root of the velocity of the gases along the tube, while Werner, 
Halliday, Carcanagues and Brille show that it depends on the 
velocity. In 1897 T. E. Stanton investigated the " Passage of 
Heat between Metal Surf aces and Liquids in Contact with Them " 
(Phil. Trans. Roy-Soc, Vol. cxc A, p. 67) and showed that the 
transmission of heat varied with the velocity of water, and that 
although not stated in the paper in words, the formula of Rey- 
nolds is applicable to the liquid side as well as the gaseous side of 
a surface. 

Prof. John T. Nicolson (Junior Institution of Engineers, 
Jan. 14, 1909, and London Engineering, Feb. 5, 1909, p. 194) 
and H. P. Jordan (Institution of Mech. Eng. of Great Britain, 
Dec. 1909, p. 1317) have each proposed formulae for the 
transmission of heat which are somewhat similar to that of Rey- 
nolds, but in these as in the works of Stanton the coefficient of 
transmission depended on the temperature difference. 

In general, the formula for the transmission of heat through 
a surface is 

Q = K(h - U)F 



HEAT TRANSMISSION 81 

where Q = B.t.u. transmitted per hour 

ti = high temperature of fluid on one side in degrees F. 
U = low temperature of fluid on other side in degrees F. 
F = surface in square feet 

K = coefficient of transmission in B.t.u. per hour per 
square foot per degree F. 

According to Reynolds 

M 
K = A + By, ori+ Bmw (11') 

According to Perry 

K = K"^, or K"mw (11") 

or later K = K ,n ™ (11'") 

Finally it has been shown experimentally that K varies with 
the temperature difference. In the above three values of K it 
is seen that for a given tube with a given discharge the value 

M 

of K would be constant, since mw = y. 

This part does not depend on the variation of temperature 
along the pipe. The quantities A, B, K" and K" f may however 
depend on temperature. In general the temperature along a 
surface changes from point to point so that t\ — t 2 is a varying 
quantity and to apply the values of K determined by experiment 
it will be necessary to find what is the mean difference in tem- 
perature if the temperatures of the substance at the ends of the 
surface are given. The formula to be used for heat transfer is 
then 

Q = (K for mean At) (Mean At)F (12) 



MEAN TEMPERATURE DIFFERENCE 

The question then arises: What is the value of mean Ml 

K f 

There is evidence to show that K = ( .. • To determine an ex- 
pression for mean At for any value of n, two cases will have to be 
considered, one for any value of n except zero and the other for 

n = 0. The reason for this is the fact that I x n 'dx = , , i x n ' +1 

J *">+ 1 

for all values of n' except — 1. Forn' = — 1 the value of the 



82 



HEAT ENGINEERING 



integral becomes log x. To find the values of mean At the 
two cases are considered. 
First case: 

K = constant, or n = .0 

dQ = K(t hx - t rx )dF = - 3600 M h c h dt h = ± 3W0M c c c dt c (13) 
dQ = heat per hour in B.t.u. for surface of dF 

sq. ft. 
K = heat per hour in B.t.u. per square foot per 

degree 
hx = temperature in degrees F. in warm sub- 
stance at point x 
t cx = temperature in degrees F. in cool sub- 
stance at point x 
M c and M h = weight of substances flowing per second 
c c and c h = specific heat of substances in B.t.u. per 
pound per degree F. 



Temperature of 

Cool Substance 



Transmission^- 
Surface 



fl * * Cool 

Substance 




Temperature of 
Warm Substance 

Fig. 26.— Parallel flow. 



The minus sign before M h is used because dt h is negative as 
dF increases, measuring from the inlet end of the warm sub- 
stance, where the temperature is ti&, toward the outlet where the 
temperature is £ 2 &. 

If the warm and cool substances flow in the same direction the 
arrangement is called a parallel flow arrangement (Fig. 26) and 
the temperature of the cold substance increases with dF hence 
dt c is positive. With the warm substance flowing in the opposite 
direction from the flow of the cool substance the arrangement is 
called counter flow (Fig. 27). A minus sign is required before 
the dt c as the temperature decreases with an increase of dF. 

If ti c is the temperature of the cool substance at the end cor- 
responding to the entrance of the warm substance and t 2c is that 
of the cool substance at the point of exit of the warm substance, 
the following is true. 

M h c h (t lh - t 2h ) = T M c c c (t lc - t^) (14) 



HEAT TRANSMISSION 



83 



The upper (minus) sign is for parallel flow and the lower 
(plus) sign is for counter flow in this equation. This of course 
assumes that all the heat leaving the warm substance goes into 
the cool substance and hence there is no radiation loss. 

The following notation may be used: 

hh ~ he = Ati 

hh — he = A^2 



and from (13) it is seen that 

M h c h (t lh - t xh ) = + M c c c (ti c - t xc ) 
± M h c h 



or 



txr. — 



M c c c 



ft 



txh) ~f" he 



At x = L 



* _ f _L MhCh / X MhCh + 

t X c — t x h Jl n/r _ txh "•" 71 /T _ t\h 



M c c c 



M c c c 



(15) 




Temperature of 

Warm Su-b.stan.ee 

-Counter current flow. 



Now 



Fig. 27 

^ = l 1 ± J^J 

d Atx 
M h c h 
M c c c 



clL 



:.dt 



xh 



1 ± 



Substituting this in (13) gives 



KAt x dF = T7 — dAh 



1 ± 



M h Ch 



dF = 



3600 



McCc 
M h c h dAt, 



K M h c h At, 



3600 
K 



~ M c Cc 

M h c h ' 

± M c c c 



(16) 



(17) 



Aii 



84 



HEAT ENGINEERING 



Now H = i£(mean At)F = 3600M h c h (h - t 2 ) 

A , 3600ilf^fti - h ) 
meanA£ = ™ 

Substituting for F its value from (17) 

M h c h (h - h) 



meanA£ = 



M h c h 



1 + 



_i *h 

M h c h 10g At 2 



(18) 



~ McCc 



From (14) 



, Mhc h _ 



tic — tic 



- M c Cc 



tlh 



tlh ■ 

' tic 



■ (tlh — t2c) 



.'. Equation (18) reduces to 



mean At 



tih 
Ah - AU 
tm — foh 

Ah — At* 



hh 



log, 



At, 
AU 



(18') 



(19) 



This is independent of the direction of flow. The form of the 
expression is the same for parallel or counter current flow. It 



tl* 


\*h 


t*h 




y x c 


h c 


tr c 









Counter Current Parallel Current Constant Warm Temperature 

tc Outlet .> th Outlet £ c Outlet <£^ Outlet tc Outlet <t h Inle t 

Fig. 28. — Temperature range for various arrangements. 

will be found that for given temperature ranges the value of 
mean At will be greater for counter current flow. If T h is the same 
for all of the surface, Ti h and T^h are the same, and are so used 
in finding the At's. If the lower temperature, as in the case Of a 
boiler, is constant this formula takes care of that condition. 
If, however, both temperatures are constant as in the case of an 
evaporator then this formula reduces to % which of course has 
the true value T h — jT c . The temperature ranges are shown 
in Fig. 28. 



HEAT TRANSMISSION 85 

With this value of mean At determined by the differences of 
temperature the value of Q for a given F could be determined as 
soon as K is known or F could be found for a given Q. 

Q = K X mean At X F (9) 

This of course is true for the case in which K does not vary 
with At x . The equation could be used to determine K if Q, F, 
and mean At are measured. 

K f 

Second case: K = tttt^ 

{f\tx) 

dQ = — -y n AtJF = - 3e>00M h c h dt xh = ± 3600 M c c c dt xc (20) 
By the method used in reducing (17) the following results 

SQ00M h Ch 




\ " McCc ) 



r (At x ) n ' l dAt x (21) 



F = 



K'll ± 

mMhCh (At,) n - (At 2 y 



K'll ± M - hCh \ n (22) 



M c c c J 



Now 



F Uoo, A^n (mean At) = SQ00M h c h (tih - t 2h ) 
^mean isij 

. ., Z600M h c h (U h - t 2h ) 
(mean At) l ~ n = y^, 

Substituting from (22) and using (18') 

M h Ch(jtih — t 2h ) 



(mean At) 1 ~ n = 



M h c h (At^-iAt,)" 



Ati — At 2 n 

t\h — t 2 h 



mean 



[ n(Ah-Ah) li-n (23) 

^-[(A^-iAt^i 

Q - FX'(mean A*)- - "[ fflgjj (W 

This equation gives the value of Q or F if the other is known in 
terms of K' which is not the transmission per degree per square 
foot per hour, but the constant which when multiplied by 
(mean At)~ n gives the coefficient of transmission. The value of 
K' in (24) as in (12) may vary with any other quantities than 
temperature and if the value of K' for the conditions of the given 



86 



HEAT ENGINEERING 



problem as to velocity, hydraulic radius or density be substi- 
tuted the expression will be correct. 

It is well to notice at this point that ~- — gives the value 

of mean A^ within 3^2 of 1 per cent, for any value of n if Ati — AZ 2 

is less than 34 o of ~ The formulae 



or 



mean At = 



mean At = 



[ n(Ati — At 2 ) 1 1- 
(A*i)»- (A* 2 )»J 



Ati — At 2 



log. 



A^ 
AU 



need not be employed to find mean At if the change in At is slight. 
If however Ati — At 2 is large compared with the mean At then the 
formulae must be used. Having mean At for any given problem 
the heat or surface may be found by 




or 



Q = K' X (mean At) l ~ n X F 
Q = K X meanAZ X F 



depending on whether or not K varies with 
i ..the temperature. 

, DETERMINATION OF K 

The method of finding this K will now be 
considered. 
Fig. 29. — Trans- In Fig. 29 let the various temperatures at 

mission through ^ e di v i s i on lines between the various layers 
various layers. TT TT T \. 

be ti , tit t 2 and t 2 while h represents the 

temperature of the gas and t 2 represents the temperature of the 

water. The following results for 1 sq. ft. area: 

Q = K(h - h) 



- — (t n 






= y(ti - h) 



HEAT TRANSMISSION 87 

where c 1 = coefficient of heat conduction for gas film. 

c 11 = coefficient of heat conduction for soot. 
c 111 = coefficient of heat conduction for heating 

surface. 
c lv = coefficient of heat conduction for scale. 
c v = coefficient of heat conduction for water film . 

I 1 , l n , l lu , l w , l y are the thicknesses in feet of the gas film, soot, 
heating surface, scale, and water film respectively. 

Now Q — = h - h l 

c 

7" 

Q^-*. , -«i n 



III 



Q — = 'i a 

Cm 

7IV 

7 V 



Adding these Q 



-m, r^~r c v| - *i 



c in 



Q ~ 7i tTi /in ti^ rT (*i - t 2 ) (25) 

— + — +—+—+ — 
c 1 c 11 c m C IV c v 



but Q = K[h-t 



Hence X (for combination) = Tl ^ ^ ^ ^- (26) 

c i ^ C n ^ c m -r cI v^ c v 

To apply this formula the thickness of the gas film I 1 and of 
the water film V must be known. These of course depend on the 
velocity. The value of c 1 for the gas and c v for the water depend 
on the temperature of these and probably on their viscosity and 
on the hydraulic radius of the pipe. The values of the c's for 
the solid materials vary with the temperature and hence the co- 
efficient of conduction K for the combination will depend on the 
temperature. 

In the case of the coefficient of transmission for walls and 
partitions of buildings, the air film usually depends on the 



88 HEAT ENGINEERING 

exposure and kind of surface hence for this formula, when applied 

c 1 
to building calculations, the terms for the two films, y % (ti — fi) 

c v 
and j v (Z x 2 — t 2 ), are replaced by aifa — fi) and a 2 (t\ — f 2 ). 

The quantities a are given by empirical formulae which depend 

on the variable factors, y is replaced by a. K becomes 

i 
K = 



1 + ^+^,^ + 1 (27) 

01 c 11 ^ c m ^ c IV ^ a 2 

The values of K for walls and partitions in most building 
problems are used as independent of temperature since the varia- 
tion in temperature is not great in these cases. 

c 1 c v 
To determine the value j x ^ , a\ or a 2 is difficult in the case of 

transmission from gas to water, as in the case of boilers or inter- 
coolers, from water to water in the case of sterilizers or from water 
to gas as in the case of indirect hot- water heaters. These terms 
are the largest and most effective terms entering into the value 
of K and for that reason the drop of temperature in the plate 
has been assumed to be zero. The coefficients at each surface 
have been determined and this determination has been made 
experimentally, putting the heat into the form 

Q = K g (h - 6) = K w (d - U) (28) 

where is the mean temperature of the heating surface. 

Kg is the coefficient of transmission on the gas side = 

c 1 

yj = B.t.u. per square foot per hour per degree. 

K w is the coefficient of transmission on the water side 

_ 5! 

" V 

U = temperature of gas, t 2 = temperature of water. 

The values of these K's may depend on velocity, temperature 
or anything else but they are determined by experiment. 

If it is desired to eliminate the temperature of the surface, this 
may be done by finding the value of K in 

Q = K(h -t 2 )F 



HEAT TRANSMISSION 89 

K = __J (29) 

for which J_ , 1 

Kg K W 

This assumes that the drop in the heating surface and its scale 
is zero or is included in the drop in each film. 

If the expression for K is assumed to be that of Perry, 

M 
K g = K" g m g w a = K" g ^ 

r w 

These are the same as the values of Reynolds in which A has 
been made zero and B has been made K" . 

Nicolson suggests that 3 be used for K" for gases and 6 be 
used for the K" for water. These values are independent of the 
temperature and using them in (29) the following results: 

1 * +- I - = ^-+ l 

r 71/T o„„„ „.. \ 



K 3 ^£ q¥j!L 3 m o w o §rn w w w (29') 

Fg F W 

3m g Wg 
m w w w \ 1 



nigWg \ Qm w w w 



m w w 



XV <a> w 



I f «*!* =L _ M^ m 

m w w w M w Fg v 

J. / 2 + L \ 1 / 2 + L \ L 

K \ L / Qm w w w \ L I 6m g w g 

K = \2+XJ ® mwWw = 2TZ (31) 

= (ul) ~K = (2+X/ jf (32) 

In (31) or (32) K is expressed in terms of water or gas condi- 
tions and of the quantity L which is the ratio of conditions of flow 
of gas to that of water. 

It will be remembered that for a perfect gas m = ^ and this 

may be used to find m for any gas. 



90 HEAT ENGINEERING 

For different values of L the values of K are given by the table 
below. 

L K 



2.0 


3.00 


m w w w or 1.50 m g w g 


1.0 


2.00 


m w w w or 2.00 m g w g 


0.5 


1.20 


m w w w or 2.40 m g w g 


0.2 


0.55 


m w w w or 2.73 m g w g 


0.1 


0.29 


m w w w or 2.86 m g w g 


0.01 


0.03 


m w w w or 2.99 m g w g 


0.0 


0.0 


m w w w or 3.00 m g w g 



Since air or flue gas at 580° F. weighs about 3^5 lb. per cubic 
foot and a common velocity is 100 ft. per second and hot water 
which weighs about 60 lb. per cubic foot may move with a ve- 
locity of 12 ft. per second, a common value of L will be given by 

. L _ -M*. _ Jfc£W0_ 1 =0 . 006 

m w w w 60 X 12 180 

and K = 3.00 m g w g (33) 

Since K is assumed independent of t, 

Q = ^.00 m g w g Ml ~^ 2 F (34) 

Before leaving this formula it may be well to point out that 
Nicolson also suggests bringing into this expression the different 
diameters of the surface in contact with the gas and the water. 

Thus dQ —K g m g w g {ti — 6)ird g dx = K w m w w w (d — t 2 )Trd w dx (35) 

6 — t 2 K g m g w g d g 



ti — 6 K w m w w w di 



U (36) 



o u u 

h-9 = Y^L f{jtl ~ k) (37) 

e ~ ** = y^T ^ " h) (38) 

If K g m g w g is known it may be used for the determination of 
the heat, of V and {t x — t 2 ) or their mean values are known. 



HEAT TRANSMISSION 91 

Thus Q = KgmgW g 1 , j, mean (h —'t 2 )F 

This can be done for the water side 

1 



Hence K = K g m g w Q 

T , 1 T d g 
since L = -^L -j- 

2 a w 



1+L' 



and K g = 3 according to Nicolson. 

j 
This result is similar to equation (31) except for the term -f-. 

The equations (37) and (38) enable one to compute the heat 
transfer from either the water or the gas side. 

The value of 2.75 for the coefficient of (33) is recommended by 
Nicolson. 

The values of K g = 3 for gas have been determined by Prof. 
Nicolson from seven sets of experiments in which the tempera- 
tures of the water and air were varied. This variation however 
was not sufficient to show the variation of K g with temperature. 
The values obtained varied from 2.84 to 3.11 but Nicolson did 
not attempt to show the variation of this with temperature. 
The value of K w = 6 was determined from Stanton's experiments 
in which water only was used. In this Stanton points out that 
K w does vary with the temperature. Nicolson proceeds further 
to show that K varies with temperature, using superheated 
steam to transmit heat to air or water and using heated air to 
transmit heat to water. In addition to these experiments 
Nicolson discussed certain boiler tests to determine constants 
for a formula. He uses the formula 

Q = K(T - 0)F (39) 

and shows that 

K = [io+iM 1 +i) m ^} (4°) 

Q = B.t.u. per hour. 

T = mean temperature of gas along flue in degrees F. 
= mean temperature of flue in degrees F. = tem- 
perature of steam. 

* = \{T + e). 



92 HEAT ENGINEERING 

di = hydraulic mean depth of flue in inches, 
area _ diam. 
perimeter 4 

m g = weight of 1 cu. ft. of gas in pounds. 
w g = velocity of gas in feet per second. 
F = area in square feet of surface on gas side. 
M = total weight per second. 

M 

m g w g = y 

This formula is worked out from experiment. It is correct in 
form if K is of the form 

K = AAt + BAt v \ 
Q will be of the form 

Q = F[AAhAt 2 + %B(Ati A + At/ 2 ) At i /2 At/ 2 ] 

or since A£iA£ 2 may be written as At 2 and y^iAt/ 2 + At/ 2 ) may be 
written as At V2 . This reduced to 

Q = A[At + %BAt V2 ]AtF 

This is the form of (39) with the value of K substituted when 
the value of A£ is inserted. 

These are the same and hence the formula is correct in form. 
If the constants are worked out to fit certain conditions, it will 
lead to correct results when applied to such conditions. 

H. P. Jordan in the Transactions of the Institution of Mechan- 
ical Engineers for Dec, 1909, p. 1317 gives the formula in the 
form 



Q = 3600 [ 0.0015 + [o. 000506 - 0. 00045di 

+ 0.00000165 (^Y 6 -)]^} (T ~ 0)F (41) 



This is also seen to be of correct form as is the case of Nicolson's 
formula. For the range of the experiments the results are 
usable. Unfortunately these two formulae do not yield the 
same results for various problems. For instance, using a problem 
given by Nicolson the differences may be seen. 

In a Lancashire boiler the flue is 36 in.; 400 lbs. of coal are 
burned per hour with 24 lbs. of air per pound of coal. The 
gases leave the fire at 2200° F. and at the end of the flue they 
are 900° F. The steam temperature is 350° F. 



HEAT TRANSMISSION 93 

r= 2200±900 = 155()OF 

= 350° F. assumed 

Q 

n = Jr = 7 

M, 400 X 25 
m g u>v - F& - 3600 x 7 0.4 

dt = -£ = 9 
By (40): 

X = [^ + ^ V950 X 1.11 X 0.4] = 4.75 + 0.34 = 5.09 

By (41): 



K = 3600 f 0.0015 + [0.000506 - 0.00045 X 9 + 



0.00000165(950)]0.4} = 2.64 

Using the method first proposed by Nicolson and assuming a 
velocity of 3^2 ft. per second for the water, 

B ^ 3 °) : L =i = 60YK = a013 

B ^ 32 ) : K = (2+loi3) a4 = L2 

This third method is evidently incorrect due to the temperature 
effect and there is a great discrepancy between the results from 
the formula of Nicolson and that of Jordan. 

If these had been used for a mean temperature of 950° but 
with 4 in. diameter 

By (40): K = 4.75 + 0.61 = 5.36. 

By (41): K = 7.5 

rp 1 a 

For — „ — = 300 and for a number of 4-in. pipes of same area 
the results would have been as follows: 



94 HEAT ENGINEERING 

By (40): K = ~j + ± V300 X 2 X 0.4 

= 1.5 + 0.346 = 1.8 
By (41): 

K = 3600(0.0015 + [0.000506 - 0.00045 X 1 + 

0.00000165 X 300] 04}. 

= 3600 [0.00172] = 6.2 

There is no change in the third method, K being 1.2 as before. 
This is more nearly the value 1.8 which is for temperatures 
near the values used in the experiments. The Jordan formula is 
reduced from a large number of experiments and in none of 
them does K fall below 4.5. 

The formula of Nicolson does not agree with that of Jordan 
and apparently the reason for this may be said to be due to the 
fact that Nicolson' s formula has been derived for large flues 
with high temperatures and Jordan's has been derived from small 
flues and for mean temperatures of 300 °. Hence it would 
be well to restrict these to the conditions from which they were 
derived. The simpler formula for K is evidently in error on 
account of the effect of temperature. The wide variations make 
it necessary to compare the results to be obtained from other 
experiments. 

H. Kreisinger and W. T. Ray in Bulletin 18 of the Bureau of 
Mines, U. S. Dept. of Interior, on the Transmission of Heat into 
Steam Boilers, have shown that the quantity of heat from hot 
gas to hot water varies directly with the velocity and with the 
temperature to some power. The efficiency of the heat trans- 
mission increases with the decrease of diameter of tube. 

Wilhelm Nusselt (Mit. tiber Forschungsarbeiten, Heft 89) 
in his article on the transfer of heat in tubes has reduced the 
theoretical form for K with an empirical constant for the value 
of K in the formula 

Ati — At 2 



Q = KF 



, A*! 

l0g *Af 2 



K = 15.90 ^(^)°' 78 (42) 

in greater calories per hour per square meter per degree C. 



HEAT TRANSMISSION 95 

^waii = coefficient of conduction of gases at wall 
temperature, greater calories per square 
meter per hour per degree for 1 meter 
thickness. 
w = velocity in meters per second. 
C p = specific heat at constant pressure for 1 
cubic meter at condition of gas in flue. 
X = coef. of conduction of gases at mean tem- 
perature of tube. 
d = diam. of tube in meters. 

This may be thrown into a different form since 



Ub — c 

c p = spe< 
(42) then becomes 



V 

Cp = specific heat of 1 kg. of gas. 



^wall ( WPC P \ 



0.786 



To change this to K in B.t.u. per square foot per hour per degree 
the same factor 15.90 is used when the terms have the following 
meaning : 

Xvaii = coef. of conduction of gas at wall tempera- 
ture of tube in B.t.u. per hour per square 
foot per degree F for 1 ft. thickness. 
w = velocity in feet per second. 
Cp = specific heat at constant pressure for 1 cu. 
ft. of gas under conditions in tube. 
X = coef. of conduction of gas at temp, in tube. 
d = diameter in feet. 

The values of X for the different gases are given below, quoted 
from Nusselt: 

French English 

Air . 01894(1 + 228 X 10~ 5 0. . 01287[1 + 127 X 10~ 5 (* - 32)] 
C0 2 . 01213(1 + 385 X 10" 5 0. . 00814[1 + 215 X 10~ 5 (£ - 32)] 
Steam 0.0192 (1 +434X 10~ 5 0. 0.01288[l + 241 X HT 5 (*- 32)] 

Illuminating Gas 

0.0506 (1 + 300X10 -5 0. 0.03390[1 + 167 X 10~ 5 (*-32)] 

For a 4-in. pipe with 1500° F. gas and a wall at 400° F. with a 



96 HEAT ENGINEERING 

velocity of 50 ft. per second, and assuming gas to be a mean be- 
tween air and CO 2, the following results: 

X = 0.01050U + 171 X 1(T 5 (1500 - 32)] = 0.0371 
\uhzu = 0.01050H + 171 X 10~ 5 (400 - 32)] = 0.0171 

C P = 0.24 X 53 Jx 6 1960 = 0.0049 

1? _ 0-0171 / 50 X 0.0049 \ °- 786 

(%)°- 214 V 0.0371 I 

= 15.90 X 0.0171 X 1.265 X 5.55 
= 1.91 

For a 36-in. flue for the mean temperature of 1550° F. and a 
wall temperature of 350° F. with m g w g = 0.4 the following 
results : 

w g m g c p = w g Cp 

\ wa ii = 0.0105[1 + 171 X 10~ 5 X 318] = 0.0162 
= 0.0105U + 171 X 10" 5 X 1518] = 0.0377 

xr 1 K ™ Q - 0162 /0.24 X 0.4\ o-786 
X=15 - 90 "3^^l 0.0377 ) 

For a 4-in. pipe for 1550° F. for gas and 350° F. for wall with 
w g m g = 0.4, the result is 

K - 1 * on aQ162 / Q»24 X 0.4 \ °- 786 

~ 1£> - yU (1^)0.214^ Qm77 ) 

= 15.90 X 0.0162 X 1.266 X 2.2 = 0.716 

These do not check with results of Nicolson nor Jordan for 
high temperatures, and of course, since based on experiments 
in which the temperature did not rise to such high values it 
should not be used in such cases. The formula has been com- 
puted for values of A* of about 20° C. to 80° C. 

In 1914, F. E. McMullen performed a large number of experi- 
ments in the Mechanical Engineering Laboratory of the Rensselaer 
Polytechnic Institute on the transmission of heat from hot air 
to water through brass and steel pipes. The velocity of the water 
was varied from 1 to 8 ft. per second, that of the air from 2 to 17 
ft. per second. The brass pipes were of %-in. and %-in. outside 
diameter and the steel pipe was of %-in. diameter. The mean 



HEAT TRANSMISSION 97 

temperature difference was as much as 300° F. The results of 
this series of experiments showed that K varied inversely as At** 
and directly with the density of the air and the velocities of the 
water and gas. The formula becomes 

Q = X(mean At)F (43) 

_ 126mQ a - 1.75)°-W« (AA , 

K ~ (mean A*)* (44j 

or Q = K'(mean At) H F 

K' = 126mK - 1.75)°"W* (45) 

m = weight of 1 cu. ft. of air at mean temperature in 

pounds = ^ 

w a = velocity of air in feet per second. 
w w = velocity of water in feet per second. 

, AA / HlMi-At 2 ] \* Afc + Afe fAa . 

(mean A*) = ( A ^_ A ^ j OT 2~" (46) 

■* A, A, ^ 1 /^1+A^2\ 

if Afc-A*^^ — 2~~; 

From the above discussion of problems it is suggested that 
Nicolson's second formula be used for boiler problems; that 
Jordan's, Nusselt's and the Rensselaer formula be used for air 
coolers and heaters where mean At is about 300° F. 



STEAM CONDENSERS 

For transmission of heat through condenser tubes a reference is 
made to the work of G. A. Orrok, A. S. M. E. Transactions, 
Vol. xxxii, p. 1138. In this he shows that the constants of heat 
transmission from the steam to the water depends on the tem- 
perature difference to the % power, on the square root of the 
velocity of the water, on the square of the steam richness, on the 
material and its cleanness. In his formula 

Q = X(mean At)F = X"(mean At) 7/i F (47) 



98 HEAT ENGINEERING 

Q= B.t.u. per hour. 

F = square feet of surface. 
K = B.t.u. per square foot per hour per degree. 
K' = 630. 

a = cleanness factor, 1 to 0.5. 

p = steam richness factor = — f where p s is the pressure 

corresponding to temperature of steam in condenser 
and p t = actual pressure in condenser measured 
in any units. 
w w = velocity of water in feet per second. 
fi = materials factor = 1.00 for copper, 0.98 for admiralty 
metal, 0.87 for aluminum bronze, 0.80 for cupro 
nickel, 0.79 for tin, 0.75 for monel metal, and 0.74 
for Shelby steel. 

The student is also referred to Orrok's article for a bibliography 
of heat transmission for condensers. 

AMMONIA CONDENSERS 

In regard to ammonia condensers this formula should apply, 
but experimental results reported in the "Transactions of the 
American Society of Refrigerating Engineers for 1907 seem to 
indicate a much lower result. This is given by a curve in which 
K for double-pipe condensers is given by 

K = 130VW (50) 

w w = velocity of water in feet per second 

and Q = i£(mean A^)^ 

(mean An = — rr- 

T AZi 

BRINE COILS 

For brine cooling in double pipes the value of K seems to be 
given by 

K = 84w 6 

EVAPORATORS AND FEED -WATER HEATERS 

The formula of Orrok should be applicable to the heating and 
boiling of water by steam as in evaporators and feed-water 



HEAT TRANSMISSION 99 

heaters. The constants and values have been determined for 
various steam pressures and for that reason they should be ap- 
plicable. However this may be, two formulae recommended by 
Hausbrand will also be mentioned. 

One of the formulae quoted is that due to Mollier from Hage- 
mann's experiments reduced to English units: 

K = 10 + { 110 + 0.6 (t s + M~) }V^ (51) 

t s = temperature of steam in degrees F. 

t\ = temperature of liquid at entrance in degrees F. 

£2 = temperature of liquid at exit in degrees F. 

The formula which Hausbrand recommends for the transmis- 
sion of heat from steam to water which is boiling is that due 
to Jelinek's experiments. When reduced to English units it 
becomes 

d = diameter of tube in feet. 
I = length of tube in feet. 

In the above formula the material is copper and steam is on 
the inside. This formula may be correct for such apparatus in 
which there is a low velocity of the liquid. This is the only 
reason for such a formula applying, as all experiments show 
that the heat varies with the velocity of the liquid. For wrought 
iron 0.75 of the above values are to be used, 0.5 for cast iron and 
0.45 for lead. In evaporators with dilute solutions of 15 per 
cent, solid matter in solution the transmission is decreased by 
about 15 per cent, of the above for clean water while, with thick 
viscous liquids, K is about one-third of that given above. 



HEAT FROM LIQUIDS TO LIQUIDS 

The coefficient of transmission from a liquid to a liquid may be 
found from the formula (29 ; ) given by Nicolson in which the 
constant 6 is used in each term. 

K Qm w w w Qm w w w 3m w w w ^ ' 



100 HEAT ENGINEERING 

Hausbrand gives the following equation for the coefficient 
from one liquid to another through brass and copper walls. 

300 



K = 



7= + - — ^ 



1 + Ww x 1 + W 



Wi 



This refers to greater calories per square meter per degree C for 
velocities in meters per second and was calculated by Mollier 
from Joule's researches. Hausbrand suggests that 66 per cent, 
of the value of K be used for practical purposes. To change this 
to English units the following calculations are made where K f 
refers to French units and K e to English: 



Kf = K e (^)\lx^j^=^K e 



K 



'39.37\ 2 _9_ 1_ 

2.2 X % 

vy 12 We 

Wf -W e X 39>37 - 3>28 

300 
4.9 



+ 



1 + 6 \OI 1+& W. 



w 2 
28 
61.2 



+ 



1 



1 +3.31vV 1 + 3.31 
60 



w 2 

(55) 



These two formula3 do not give the same results and the 
later one is recommended for general use. 

FACTOR OF SAFETY 

In all of the above expressions for transmission it would be well 
to reduce the quantity by 34 to have an excess of surface, or what 
would be the same thing increase the surface by 33 per cent, in 
finding the area required in a problem. 

RADIATORS 

The coefficient of transmission for direct radiators for heating 
varies with conditions of the surface and the type of radiators. 



HEAT TRANSMISSION 101 

Carpenter quotes tests varying from 1.23 to 1.97 while Rietschel 
gives values of K varying from 0.51 to 2.65. A value of K 
of 1.8 will be used as an average for the computation of the 
heat transmitted for all types of direct radiators. This gives 
for such 

Q = 1.8(*. - QF (56) 

Q = B.t.u. per hour. 

F = square feet of surface. 

t s = temperature of steam or water in degrees F. 

t a = temperature of room in degrees F. 

For indirect heaters the formula 

Q = K(t s - t ^)F 

may be used in which 

K = 2 + 1.75\/^. 

w a = velocity of air across surface in feet per second. 

t 1 = temperature of air at inlet in degrees F. 

t 2 = temperature of air at outlet in degrees F. 

t s = temperature of steam in degrees F. 

Experiments have been made and plotted for the determination 
of the transmission of heat and the resultant air temperatures 
obtained with coils and cast-iron radiators and these are to 
be resorted to rather than formulae because the resultant tem- 
perature is dependent on the velocity and the temperature of 
the entering air and steam. Since in most problems low pres- 
sure steam is used the curves constructed from data of the 
American Radiator Co. and of the Buffalo Forge Co. are given, 
the first for a cast-iron pin radiator known as a Vento Heater 
and the second for the sections of pipe coils, each section being 
four pipes deep. The curves of Figs. 30 and 32 give the result- 
ant air temperatures when air at zero degrees enters the indirect 
heater and passes over one, two, three, four, five, six, seven or 
eight sections at different velocities. The velocities are found by 
finding the volume of the air at 70° F. and then computing the 
velocity by dividing this volume by the clear area through 
which the air passes. Thus the velocity to be used for the 
curve is not the actual velocity at entrance or exit but is that 
occurring when the temperature of the air is 70° F. 

The curves of Figs. 31 and 33 are those giving the average B.t.u. 



102 



HEAT ENGINEERING 



per square or lineal foot of the total heater under the conditions 
given. It will be seen that as the velocity increases the final 

240° 

220' 
200 

180* 



2 w 



bn 


160 


c 




Is 


140 


0J 




*1 




?, 


120 


3 


c 


% 


100 


V 




I 


80 


60° 




40 c 






20 



k 


































^ 


































\% 


^ 
































IY\ 


vr 
































\v 
































\ s 


\ 






























I \ 





































































































































































200 



400 



1400 



1600 



1800 



500 800 1000 1200 

Velocity in Ft. per Minute 
Fig. 30. — Temperature of air leaving Vento Heater with air at 0° F. and veloc- 
ity at 70° F. Steam at 5 lbs. gauge pressure. 

2700 
2600 
2400 



2200 



2000 
1800 



1600 



1400 



? 1200 



1000 



ffl 800 



600 



400 
200 





















































































































































































.-^ 




















































































































'/, 


'# 


% 


/. 


























'/, 






































^ 


p 






























Sy^ 


'& 


'S 





































































































5 v 

7 



200 



1400 



1600 



1800 



00 600 800 1000 1200 

Velocity in Ft. per Minute 
Fig. 31. — Heat transmitted per sq. ft. of Vento Heater with 5 lbs. steam and 
air entering at 0° F. Velocity at 70° F. 

temperature of the air falls but that the heat transmitted per 
square foot increases, due to the lower final temperature and the 



HEAT TRANSMISSION 



103 



Z4U 

220° 

200° 

180° 

„ 160° 

8 140 

+i ° 
rt 120 

S 

*-i o 

5 loo 

a 80 

s 

v o 
H 60 

o 






































fe 




































1 




































\\ 


\\ 
















• 


















v 


^\ 


\N 
































\ 


V 


s \ 


s 






























\ 


^ 




































\ 


\ 


































\ 






































































40 

o 

20 

n 











































































200 



400 



1400 



1600 



1800 



600 800 1000 1200 

Velocity in.Ft. per Minute 
Fig. 32. — Temperature of outlet from sectional heaters of four coils 
each, with air entering at 0° F. Steam at 5 lbs. gauge pressure. Velocity 
at 70° F. 



1200 
1100 
1000 
900 
800 
700 
600 
500 
400 
300 
200 
100 



































































































































































































































s 


g 


i. 


ik 


^> 






























Ss 


^ 


^ 


























4 


^ 


g 


p 


*• 


























/k 


& 


p 


*> 

































































































1 

2 
3 

4 S 

5 .2 

6 tJ 
7£ 
8 



200 



400 



1200 



1400 



1600 



1800 



600 800 1000 

Velocity in Ft. per Minute 
Fig. 33. — Heat transmitted per lineal foot of 1 in. pipe for 4 row sections 
of coil heaters with 5 lbs. steam and air entering at 0° F. Velocity' at 
70° F. 



104 HEAT ENGINEERING 

higher velocity. As the number of sections is increased the final 
temperature rises by decreasing increments since the tempera- 
ture difference between the steam and air is less for the succes- 
sive sections, decreasing the heat transmitted, and the heat per 
square foot is less for the same reason. 

Since air often is supplied at a higher temperature than 0° F. 
it is well to understand how such curves may be used for these 
conditions. Suppose air at 40° F. and 600 ft. velocity is to be 
heated in Vento heaters. It will be seen that one section would 
have been required to give this temperature and if a temperature 
of 100° F. had been desired three sections would be required to do 
this with zero air. The first section would bring the air to 
40° F. so that only two sections will be required for 40° air. 
Now with zero air one section at 600 ft. velocity will transmit 
1450 B.t.u. per hour per square foot and three sections will 
transmit 1150 B.t.u. per square foot per hour. The amount 
transmitted by the last two sections will therefore be 

1150 X 3 - 1450 = 2000 
or 9 = 1000 B.t.u. per square foot. 

If two sections had 0° F. air supplied the curves show that 
1275 B.t.u. per hour per square foot would have been trans- 
mitted but in this case, the entering air being at 40° F., there 
is a smaller At and hence a smaller transmission per square foot 
per hour. 

These curves can be used with certainty as they are the 
results of experiment and they displace any formula for compu- 
tations. They are better in this case as they show what may 
be obtained actually in practice. 

HEAT THROUGH WALLS AND PARTITIONS 

The heat transmitted through walls and partitions is com- 
puted by 

Q = K(h - t 2 )F (58) 

K= 1 

where ' 1 V_ T 1 (59) 

a x "•" c' + c" + • * ' a 2 

Q = B.t.u. per hour. 

h = temperature on one side in degrees F. 
U = temperature on other side in degrees F. 
F = area in square feet. 



HEAT TRANSMISSION 105 

The value of a is found by experiment in the form 

a = d + e+ y - ^ 0Q (60) 

d = constant depending on condition of air. 

= 0.82 air at rest in rooms or channels. 

= 1.03 air with slow motions, as over windows. 

= 1.23 air with quick motions, as outside of build- 
ing. 
e = constant depending on material. 
T = temperature difference of film of air at wall. 

Values of e 

Cast iron . 65 

Cotton and fabrics 0. 65 

Charcoal 0. 71 

Glass 0.60 

Metal, polished 0.05 

Masonry . 74 

Paper 0.78 

Rusted iron . . 69 

Water 1 . 07 

Wetted glass 1 .09 

Wood . 74 

For masonry walls, of thickness I ft. 

T = 16.2 - 4.00Z (61) 

For wood T = 1.8° F (62) 

For glass T = Y 2 {i x - t 2 ) m (63) 

The values of c are given below: 

Air (still) 0.03 

Brass 61.00 

Building paper . 08 

Cork 0.17 

Cotton and felt . 02 

Glass 0.54 

Masonry and plaster . 46 

Sandstone . 87 

Sawdust . 03 

Slate 0.19 

Terra cotta 0.54 

Tin 35.60 

Wood 0.12 



106 HEAT ENGINEERING 

Experiments show that the values of c change with the 
temperature, but for the temperatures used in practice the 
variation is not great. 

EFFICIENCY OF HEAT TRANSMISSIONS 

The efficiency of a heating surface is the ratio of the heat 
abstracted by the surface from the gases to the amount of heat 
which the gas could give up to the surface. If T lh is the tem- 
perature of the hot substance entering and T 2 h is the temperature 
of the hot substance leaving the tube and T c is the lowest tem- 
perature of the cool substance in contact with the opposite side 
of the surface, the expression for efficiency is 

T\h — Tin .. Tih — T c .. At 2 /t%A\ 

"= T lh -T c =l -T^rr 1 -^ m 

This is true since the heat available in M lbs. of gas of specific 
heat c is 

Qi = Mc(T lh - T e ) 

and the amount utilized is 

Q 1 -Q 2 = Mc(T lh - T 2h ) 

In a given experiment the temperatures are known but with a 
given T\h and T c and with a given surface it is necessary to find 
T 2 h in order to ascertain the efficiency. 

There are two cases to consider: 

First, if the cool substance changes in temperature along the 
surface. 

Second, if the cool substance remains at a fixed temperature as 
in the case of a boiler. 

In the first case it has been shown earlier in the chapter that 

-M h c h dt h = K(t xh - t xc )dF 

__ M h Ch 1 ,, 

— at = ^r- -77 — r ah 

JV (t x h — txc) 

_ M h c h 1 dAt _ dM 

" K Mhth At ~ COnSt - KAt 

± M c Cc 

If K is independent of temperature this becomes 

MhC h 
nvr ~ M c Cc Ah 

FK Mncn =1 ° ge A^ 



HEAT TRANSMISSION 107 

MhCh i he — he hh — t\ c — (hh — he) 
But 1 ± -tj — = 1 — : 7— = , , 

McCc hh ~ Hh hh — Hh 

__ Ah - Ah FK(Ah - Ah) . At 1 

Hence FK -^ — 7- r^r = ?\ = log e 77- 

M h c h (tih-hh) Qi & Ah 



Ah - Ah 



A^i _ -FK 

Ah 



-FK^ 



Mi 



, - 1- e- FK —QT- (66) 

This shows that the efficiency of a surface increases with the 
increase of area, with the increase of the constant K or of the 
difference of At, but decreases as Qi increases. If, however, the 
area F is increased for a fixed length of tube by increasing the 
diameter, this apparently shows that the efficiency will increase 
with the diameter. It is known, however, that K varies with the 
velocity and hence this term would decrease with the square of the 
diameter so that the product FK would decrease with the increase 
of diameter and hence the efficiency would decrease. On the 
other hand, if d is decreased for a given length, the term FK is 
increased. In the above discussion Qi and (Ah — At 2 ) have been 
assumed to be constant but this would not be so for if d were de- 
creased the velocity would be increased and with it Qi for a given 
length would be increased, but (A^i — Ah) would increase to a 
greater degree and the final result would be an increase of 
efficiency. 

If K depends on (At)~ n , the integration gives 



Qi n 




nFK' Ah n - Ah n 




Qi Ah — Ah 




Ah nFK' \A<2 / 




» ft (*r*-£^) 




Ali-»-^-Att-» 

Ah 


(67) 




In this as before the efficiency increases with FK' 


and de- 


creases with Qi. 





108 HEAT ENGINEERING 

In the second case the temperature is constant on the cold side 
and the expression then becomes 

+ M h c h dt h = K(t xh - t c )dF 

+ if dt is measured from cold to hot. 
If K is independent of the temperature this becomes 

KdF dt h 



M h c h 


Ixh *c 




K F - 

M h c h 


t Tin — 

i 2h — 


T c 

T c 


Tih — T c 
Tin — T c 


KF 

= e Much 




T=l 


KF 

— € Mhch 





In this as before the efficiency increases with the product KF 
and decreases with the quantity M. Thus if the diameter of a 
pipe is increased the velocity decreases for the same M or M 
increases for the same velocity. In each of these the exponent is 
decreased since M would vary with the square of d directly and 
K inversely. Hence the large diameter means a smaller effi- 
ciency. If M is increased for a given tube by increasing the ve- 
locity K will increase in the same ratio and the efficiency may not 
be changed. If the length is increased F is greater and the 
efficiency increases. Since this increase of efficiency is logarith- 
mic, the increase beyond a certain value of F is very slow. 

If K depends on the temperature the following results 

ff = (At Y -HM (69) 

This leads to an expression similar to the one used previously. 

Kreisinger and Ray, in Bulletin 18 of the Bureau of Mines, 
show by results of tests on an experimental boiler that the effi- 
ciency decreased with the velocity to a certain point although the 
quantity of heat increased. In these experiments the value of 
M increased more rapidly than the coefficient of conduction. 
Of course if at the same time F were increased to its proper 
amount the efficiency would increase. In these experiments it 
was shown that the absorption of heat increased with the in- 
crease of the initial temperature although the rate of increase 



HEAT TRANSMISSION 109 

of heat gradually diminished. The enlargement of the di- 
ameter of the tubes reduced the efficiency. Increasing the 
length of a tube increased the efficiency of the surface. 

For a boiler Kreisinger and Ray have used the form for the 
value of K suggested by Perry. 





K = K'mw 


Hence 


M h c h dt = K'm h w h AtdF 


it 


M h 
m h w h = -y~ 




. dt _ K> 
' * A* FoC h a 




lo ^M^Fj h F 




K ' F 

F ° Ch 

At x = Ati e 




_ K ' F * 

H Foch 



7] = 1 — € 



(70) 



This expression for efficiency does not change with change in 
weights and therefore does not give curves similar to those found 
by experiment. 

If the Reynolds form be used 



M h c h dt h = K'(A + Bj^AtdF 



this leads to _ ( A+B r) Fx 

A 1 M MhCli 

At x = Ati e 
and _ ( A+B r) F * 

i Mhch 

7} = 1 — € 



(71) 



This equation yields a curve similar to those found experi- 
mentally. 

PROBLEMS 

A number of problems will now be computed, applying the 
formulae best suited for the work. 

Problem 1. — A 4-in. boiler tube is used in a return tubular boiler with a 
grate of six times the flue area. Fifteen pounds of coal are burned per 



110 HEAT ENGINEERING 

square foot of grate per hour with 25 lbs. of air per pound of coal. The gases 
entering the tubes are 1350° F. and at exit they have been reduced to 600° F. 
What is the average value of the heating surface of tubes and how many 
pounds of water at the boiling point will be evaporated if the pressure is 
130.3 lbs. gauge? 

The formula to use in this case is that of Nicolson (40). 

The gases are of practically the same density as air, hence B = 53.35. 

Mass of gas per tube = tttX WK) = 0-0567 lbs. per second. 

1350 + 600 
Mean temperature of gas = k = 975 b . 

Temperature of steam and water in boiler = 355.8° F. 

<f> = ^(975 + 355.8) = 665.4 
Hydraulic radius = H = 1. 

Area of tube = ~ = 0.0872. 
144 



T665.4 V665.4,, 0.05671 /n „ orrox 

Q = Lw + -ir (1 + ^oosTij (975 - 355 - 8) 

= [3.327 + 0.839] (619.2) = 2570 B.t.u. 
Now ri 45 = 865.2 B.t.u. 

ok 7 A 

Hence the pounds of steam per square foot = ^ fir . ~ = 2.97 lbs. 

This result is an average value per square foot for the whole boiler. The 
value of the shell directly over the fire increases the average to more than 
this value. If the temperature entering the flue had been higher this value 
would be greater. 

For water tube boilers it is suggested that the area be considered the area 
between tubes and that the hydraulic radius be taken as the distance between 
tubes. 

Problem 2. — Thus if the gases enter the tubes at a temperature of 1900° 

M 
F. and leave at 600° F. and if the tubes are 3 in. apart and the value of — 

is the same as before, the following results would hold for a water tube boiler. 
Mean temperature gases = 1250° F. 

<f> = 803 

M 
Hydraulic radius = 3 in. ~y = 0.65 

Q = [H + ^p (1 + H)0.65] (1250 - 355.8) 

= [4.015 + 0.613] (894.2) = 4120 B.t.u. 

4120 
Weight of steam = ^ fi - - = 4.76 lbs. 

Problem 3. — Air at 350° F. is to be cooled to 75° F. in an intercooler 
made up of %-in. tubes with the air passing through the tubes and cooled 



HEAT TRANSMISSION 111 

by water entering at 60° F. and leaving at 90° F. Find mean At if (a) K 

K' 
is independent of temperature and (6) if K = , ,^ - 

In this problem it is evident that a counter-current flow must be em- 
ployed since the water leaving the intercooler is higher than the air leaving. 

A*! = 350 - 90 = 260° F. 
Ah = 75 - 60 = 15° F. 

„.„ ^ 1/Ah + At 2 \ 1/260 +15\ 137.5 
Now*** -Ah = 245 > io(— 2— j = To (~ 2~ ) = ~W 
Hence the exact methods must be used. 

, N w Ah -Ah 260-15 245 

(a) Mean At = -—^ = T^eo = 2.3 X 1.237 = 86 ' 2 

° ge At 2 ge 15 

/mat * A , r K<Afr-Ato i* r 81 - 7 i 3/ ^ o KK 

(6) Meah A* = [ ^ _ ^h J = [o^— 2:46oJ = 95 ' 5 

Problem 4. — Find the heat per square foot for temperature conditions of 
problem 3 if the water has a velocity of 5 ft. per second over the tube and 
the air has a velocity of 20 ft. per second. The pressure of the air is 70.3 lbs. 
gauge. 

350 + 75 
m for air at 85 lbs. abs. and at ~ or 212.5° F. 

85 X W4 0.341 lbs. 



53.35 X 672.5 
(a) Using (30) and (32): 

^W^° = 0.02135 

K = 2+0.022 X °' 341 X 20 = 20 - 2 
Xow Q = K (mean At) = 20.2 (86.2) = 1740 B.t.u. 

(6) Using (41): 

c r 3 

= 3600 ! 0.0015 + 0.000506 - 0.00045 X 7^ + 0.00000165- 



w ^ 16 

1 
J 



/212.5 + 75\ 1 1 

( 2 ~ ) J ( 20 X °' 341 ) } ( - 212 ' 5 ~ 75 ) = 21-6(137.5) = 2970 B.t.u. 



(c) Using (42'): 

X wa u = 0.01287(1 + 127 X 10" 5 X 43) = 0.0135 
X oas = 0.01287(1 + 127 X 10" 5 X 180.5) = 0.01583 
0.0135 r 20 X0.24 X 0.3411 0786 
(Hs) 0214 l_ 0.01583 

Q = 15.0 X 86.2 = 1293 B.t.u. 

(d) Using (43) and (44) : 



v-^ [- -7w»«sr- = 15.0. 



_ 126 X 0.341 X ( 18.25)°-«(5)K ^ a 
K ~ ' (95.5) H = 36 ' 6 

Q = 36.6 X 95.5 = 3490 B.t.u. 



112 HEAT ENGINEERING 

The results do not agree very well and result (d) is very high with (6) next, 
while result (c) is very low. The reason for the low value of (a) is due to the 
fact that the data from which equation (32) was deduced holds for smaller 
differences of temperature. Equation (32) will in general give results on the 
safe side as the temperature differences for which it holds are small. 

Problem 5. — A sterilizer operates on a counter-current principle with the 
warm water entering at 212° F. and leaving at 75° F. and the cool water 
entering at 70° F. and leaving at 202° F. The liquid moves at a velocity of 

2 ft. per second. Find the mean At : (a) if K is constant, (6) if K = ~rji^ 

and (c) arithmetic mean At of Ai's at the two ends. 

Ah = 212 - 202 = 10° F. 

Ah = 75 - 70 = 5° F. 

AU + At* 
(c) ^ 2 =7^°F. 

. . _ _ Kt Ah-At 2 10-5 

(a) Mean A* = ——- = 2XX O301 = 7 * 24 

A t 2 

(b) Mean A< = [ A ^ 2 _ A ^ 2 J = [062^^23eJ = 7 ' 3 ° 

Problem 6. — Find K and Q per square foot for problem 5. 
Use (29) with K" = 6 for water and at 150° F., m w = 61.2: 

K = — i ^ — = 3m w w w = 3 X 61.2 X 2 = 367 



6ra w iy w 6m M ,to w 

Q = 367 X 7.24 = 2654 B.t.u. 

By formula (55) X = — ^- = 171 



l + fV2 



Q = 171 X 7.24 = 1240 B.t.u. 

Problem 7. — If the walls are He in. thick in the sterilizer what is the 
drop in temperature in the copper of the tube to transmit the heat of prob- 
lem 6? 

Using equation (3) and the tabular value of C for this: 

124Q _ 239.0(1 + 0.00003X150) ft _ y 
16 X 12 

tl ~ k = 16X12X241 = ' 027 ° 

Even in this case the fall of temperature occurs mainly in the films of 
water. 

Problem 8. — The temperature of the condensing water is 70° F. at inlet 
and 80° F. at outlet. The temperature of the steam is 105° F. Find mean 
At and the heat removed per square foot of cooling surface of admiralty 



p = - = v^t^ = 0.914 



HEAT TRANSMISSION 113 

metal if the surface is clean and the pressure in the steam space is 1.2 lbs. 
absolute, and the velocity of the water is 4 ft. per second. 

Ah = 105 - 70 = 35° F. 
At 2 = 105 - 80 = 25° F. 

I~W(35- 25)19* r 1.25 l 8 /r nnn 
Mean At = |_35H-25W| = Ll.560-l.496j = 29 * 9 

p, _ 1.098 

p t " 1.200 

_, 630 X 1 X 0.914 2 X 0-98 X VI „ _ , 
K - ■ (2^9)H 673 B ' tu - 

Q = 673 X 29.9 = 2010 

Problem 9.— Steam at 215° F. is used to heat water at 60° F. to 180° F. 
with the steam inside of copper pipes 3 A in. in diameter and 80 in. long. 
The velocity for the water is 2 ft. per second. 

Using Orrok's formula : 

Ah = 215 - 60 = 155° F. 
At 2 = 215 - 180 = 35° F. 

Mean At = Ll55W - 35JiJ = Ll.878-1.56j = 82 

zr _ 630 X 1 X 1 X 1.00V2 _ 

A ~ (82)W " 51d 

Q = 513 X 82 = 42,100 B.t.u. per square foot. 

Using equation (51) of Hagemann's, 

X- 10+ jllO + 0.6(215 +H>±«9)}^ 

= 10 + 441 = 451 

155 4- 35 
Q = 451 X y — = 42,800 B.t.u. per square foot. 

Problem 10. — Suppose the pipe in problem 9 was used to boil water at 
180°, find K and Q. 

Using equation (52) of Jelinek: 

iC = _mo = = 1965 

M2V?* X80 
Q = 1965 X (215 - 180) = 68,800 B.t.u. 
Problem 11. — A brine cooler has brine with a velocity of 1 ft. on one side at 
20° to 10° F. and ammonia on the other at 0° F. Find K and Q per square 
foot. 

In this problem the method will be to use the same constants as those 
used for steam to water. Using Hagemann's equation (51): 

K = 10 + [llO + 0.6 (0 + ^p9) } VI 

= 10 + 119 = 129 
Q = 129 X 15 = 1935 B.t.u. per hour per square foot. 



114 



HEAT ENGINEERING 



For 4 ft. velocity this would give for K 

K = 10 + 238 = 248 
Problem 12. — An ammonia condenser uses steel pipes with water from 
60° to 80° F. and ammonia at 90° F. The water velocity is 4 ft. per second. 
Using equation (50) : 

K = 130 X VI = 260 

Q = 260 X o^7 10 q = 260 X 18.3 = 4750 B.t.u. 
2.6 log e 3 

Problem 13. — Find the number of sections required and the average heat 
transmitted per square foot for vento heaters to operate with steam at 220° 
F. (5 lbs. gauge) and to heat air from 60° F. to 114° F. when delivered across 
heater at 1200 ft. per minute. 

From Fig. 30: Two sections will heat zero air to 60° F. and five sections 
will heat zero air to 115° F. at 1200 ft. per minute; 5 — 2=3 sections are 
required. 

From Fig. 31: The average transmission for two sections with zero air 
is 2050 B.t.u. while with five sections 1600 B.t.u. are transmitted. The 
amount for the last three of the five sections will be 



Average transmission = 



5 X 1600 - 2 X 2050 



= 1300 B.t.u. 



To find the result of this problem by formulse in place of using the 
experimental curves the following is given: 

60+ 115 



Mean At = 220 



= 132.5 



K = 2 + 1.75 \J^~ = 9.83 
Q = 9.83 X 132.5 = 1300 B.t.u. 




Brick .. a 
Air Sp 

Fig. 34. — Section of wall for cold storage room. 



Air Space Cork 

JJoard 



Problem 14. — Find the value of K for the wall of a cold storage warehouse 
shown in Fig. 34. 

T = 16.2 - 4.00 X 3 = 4.2 



a for outside = 1.23 + 0.74 + 

= 2.28 
a for air space 



42 X 1.23+31 X0.74 
1000 



X4.2 



- 0.82 + 0.74 + * 3X0 - 82 +5 X0 - 74 X 4.2 
a for inside J 1000 

= 1.79 



K 



HEAT TRANSMISSION 115 



1 2 1 1 067 033 017 J_ 

2.28 + 0.46 + 1.79 + 1.79 + 0.46 + 0.17 + 0.46 + 1.79 



~ 0.438 + 4.35 + 0.558 + 0.558 + 1.455 + 1.94 + 0.37 + 0.558 

= 0.0978 
It will be noted that the amount of air space does not aid in the insulating 
value of the wall because the air can and will circulate. The value of this 
lies in the resistance at the two surfaces which amounts to almost as much 
as 8 in. of brickwork. The various terms of the denominator show the 
values of the various elements of the wall as heat insulators. Thus 4 in. 
of cork is as valuable as 11 in. of brickwork and may be used when space 
is of value. The wall above will transmit 2.35 B.t.u. per square foot per 
degree difference in 24 hours. For a temperature difference of 60° F. this 
amounts to 141 B.t.u. per day per square foot. 

Topics 

Topic 1. — By what methods is heat transmitted? Explain the pecul- 
iarities of each method. Give the Stefan-Boltzmann Law. For what is 
this used? How is it applied? 

Topic 2. — What is the law of conduction as stated by a formula? Is 
the coefficient of conduction a constant? What are the dimensions of this 
coefficient? What fixes the amount of heat carried by convection? Why 
is convection of value in engineering problems? Is the heat in these 
problems carried by convection or conduction? 

Topic 3. — How may it be shown that the films of substance at the surfaces 
of a partition or plate offer the greatest resistance to flow? What is the 
effect of velocity on film resistance? Explain why this is true. What was 
the work of Dalby? What was the work of Reynolds? 

Topic 4. — Prove that 

Q = K'mwit - e) 
Is this the equivalent of Reynold's expression 

Q = {A + Bmw){t - 6) 

Show that — = mw 

F 

Topic 5. — What is meant by hydraulic radius? Determine the mean 
temperature difference along a surface if K is constant. 

!*'■»» A^i — At 2 ~ 

Mean A* = — 



L 



l0g 'A«- 2 



Topic 6. — What is meant by the constant K? Prove that 

T n(Ati - At 2 ) ~| — 
Mean At = r^, 1 — tttL U-» 



if K = J 



{At)-' 
Topic 7.— Prove that 



A' 






116 HEAT ENGINEERING 

To what does this relation reduce for ordinary transmission through 
thin partitions? 

Is c a constant? What is the form to which Nicolson reduces this value of 
K above? 

Topic 8. — Given: -^ = 5 + « 

K 3m g w g om w w w 

reduce K = ~ , j 6m w w w = „ ° ~j- 

Topic 9. — What are the formulae of Nicolson, Jordan and Nusselt? Are 
these formulas applicable to the same conditions? On what does the heat 
per square foot per degree per hour depend? Give details and reasons. 

Topic 10. — For what conditions is the Rensselaer formula applicable? 
For what is Orrok's formula used? On what does the K of Orrok's formula 
depend? On what for the Rensselaer formula? 

Topic 11. — Is Orrok's formula applicable to ammonia condensers, 
evaporators and feed-water heaters? Are special formulae given for these 
forms of apparatus? Give the formulae used for finding K for these. 
What is the value of K for direct steam radiators? For indirect steam 
radiators ? 

Topic 12. — Explain the method of finding the heat transmitted per square 
foot of surface per hour in Vento heaters and in pipe coils used as indirect 
heaters. 

Topic 13. — Explain the method of finding the value of K for a wall or 
partition. 

Topic 14. — Derive the expression for the efficiency of heat transmission 

7j = 1 — e Q 

when K is independent of the temperature. Give all steps and discuss the 
variation of the efficiency of a surface with velocity, diameter of pipe, 
length, temperature, and heat. 

Topic 15. — Starting with Reynolds' value of K 

K =K'(a +B~j 

reduce the expression for the efficiency of a heating surface. 

(* + *{&* 



= J - 



6 Mhch 



PROBLEMS 

Problem 1. — A piece of glass is held at 750° F. and a shield covers two- 
thirds of this. The surface of the glass is 2 sq. ft. The surface of the 
shield is 20 sq. ft. The shield is held at 125° F. by a water-jacket. How 
much heat is removed by the water-jacket? 

Problem 2. — A 4-in. boiler tube has gas entering at one end at 1400° F. and 
leaving at 550° F. with steam at 120 lbs. gauge pressure. The coal is burned 



HEAT TRANSMISSION 117 

on a grate of eight times the area through the tubes at a rate of 15 lbs. per 
hour with 30 lbs. of air per pound of coal. Find the value of K and the 
number of B.t.u. per square foot of surface. How many square feet of 
surface would be required per boiler horse-power? (One boiler horse- 
power equals the evaporation of 34^ lbs. of water at 212° F. per hour into 
dry steam at 212° F.) 

Problem 3. — A feed-water heater uses steam at 3 lbs. gauge pressure to 
heat 6000 lbs. of water per hour from 60° to 200° F. The water is passed 
through at a velocity of 2 ft. per second. Find mean At assuming K con- 
stant. Find K by two formulae. Find the square feet of surface necessary. 
Of what material will the tube be made? Why? 

Problem 4. — An economizer is used to heat 6000 lbs. of water per hour 
from 60° F. to 200° F. by using hot gas at 500° F. in 3-in. iron flues in which 
the temperature is reduced to 350° F. The velocity of the gas is 25 ft. 
per second. What is the value of mean At? Find K by Nusselt's formula. 
Find the amount of surface required. Also use the simple formula 
suggested by Nicolson in which L is employed. 

Problem 5. — In a condenser the water enters at 50° F. and leaves at 65° 
F. with a pressure in the condenser of 0.6 lbs. absolute and a temperature of 
80° F. What is the value of mean At for this case? Find K for copper 
tubes if dirty. Find the square feet per kilowatt of turbo generator if 
steam consumption is 14 lbs. with x of exhaust steam equal to 0.95. 

Problem 6. — Find the size of the condenser for an ammonia plant to 
remove 200,000 B.t.u. per hour with water flowing at 5 ft. per second in the 
double pipe, entering at 65° F. and leaving at 80° F., when the ammonia is 
at 100° F. 

Problem 7. — A boiler using a 24-in. flue has gas entering at 1600° F. 
and leaving at 1000° F. The steam is at 300° F. The velocity of the gas 
is 100 ft. per second. Find the heat transmitted per hour per square foot 
of flue. 

Problem 8. — An intercooler receives air at 250° F. and cools it to 80° 
F. The water enters at 50° F. and leaves at 70° F. Find mean At for 

K' 
parallel and counter- current flow using K = constant and K = —jr. Find 

the heat transmitted per square foot per hour. 

Problem 9. — Find the surface required in an interchanger cooling 7000 
lbs. of water per hour from 220° F. to 80° F. by water entering at 60° F. and 
leaving at 200° F. Use K = constant. 

Problem 10. — Find the surface required to boil 500 lbs. of solution at 
200° F. by steam at 250° F. if tubes 3 ft. long and 3 in. in diameter are 
used and the heat of vaporization of the liquid is 750 B.t.u. 

Problem 11. — Assume air at 50° F. and move it with a velocity of 1200 
ft. per minute over the Vento heaters or coils to heat it to 105° F. How 
many sections will it take for the Vento heaters and for the coils? How 
many square feet of each will be required to heat 200,000 cu. ft. of air per 
hour? 

Problem 12. — Find the K for a wall composed of 16 in. of brickwork, a 
2-in. air space and 12 in. of brickwork with 1 in. of cement plaster. 



CHAPTER IV 

AIR COMPRESSORS 

Compressed air is used for many purposes for which it would 
be difficult to employ other media. For operating small tools, 
rock drills, and hoists and for the transmission of power over con- 
siderable distance it replaces steam with which the loss due to 
condensation is very excessive. The efficiency of transmission 
of power by compressed air is not equal to that of electrical 
transmission nor is it as flexible, yet in certain cases for some 
reasons it is of value. For cleaning materials with a sand blast 




Fig. 35 



-Section of radial blade fan 
blower. 



Rotary blower, section. 



and for use with the cement gun compressed air is necessary. 
To force air into a furnace of a boiler or for a blast furnace a 
compressor of some form is required. Compressed air may be 
used in air-lift pumps and in direct air pumping. In supplying 
air to divers, in the making of liquid air and in the operation 
of street cars compressed air is required. 

To compress this air several types of compressors are used. 
For low pressures up to 10 oz. per square inch above or below the 
atmosphere, centrifugal or fan blowers, Fig. 35, are used. In 
these, radial or curved blades are driven at a high speed causing a 

118 



AIR COMPRESSORS 



119 



flow of air. These are used for ventilation of buildings ; for supply 
of air to boiler furnaces or cupolas; for forges; for suction in an 
induced draft system or for the conveying of light materials. 
In forge work Sangster allows 140 cu. ft. of free air per minute 
per forge at 2 oz. pressure. If an exhaust fan is used 600 cu. ft. 
of air per minute at Y± oz. pressure are handled per forge. The 
air required in cupolas is 40,000 cu. ft. per ton of iron melted. 
For ventilation 2000 cu. ft. of air are allowed per person per hour. 

For pressures of from 8 oz. to 7 lbs. per square inch, rotary 
blowers such as that shown in Fig. 36 are used. These are run 
at a sufficient speed to get the necessary discharge in cubic feet. 

For pressures up to 35 lbs. turbo compressors of the form shown 
in Fig. 37 are used although piston compressors are used for this 
pressure at times. The air at this 
pressure is used in blast furnaces 
and converters and the compressors 
of the piston type for such are 
known as blowing engines or blow- 
ing tubs. These are only special 
forms of air compressor. 

For higher pressures than 35 lbs. 
the piston compressor is in general 
employed, although the hydraulic 
compressor or the Humphrey explo- 
sion compressor could be used. For 
high pressures it will be shown 
that efficient work necessitates the 
compression in one cylinder to a 
pressure much below that desired and then a further compres- 
sion of this air to a higher pressure. It may be that a third 
compression is used before the desired pressure is reached. Each 
one of these cylinders is known as a stage. The compressor men- 
tioned above would be known as a two-stage compressor. In 
general one stage is used to 70 or 80 lbs. gauge pressure while 
from that to 500 lbs. two stages are used, and three stages from 
500 to 1500 lbs. Above this four stages would be used. Fig. 
38 shows a two-stage compressor. In this air is sucked into the 
center of the piston A by the vacuum produced behind the piston 
when the piston moves to the left, the air flowing through an open- 
ing left at the periphery B, as shown in Fig. 39. The air on the 
left of the piston is compressed and after it reaches the pressure 




Fig. 37. — Section of turbo blower. 



120 



HEAT ENGINEERING 



existing above the mushroom valve at C this opens and allows 
the air to exhaust into the discharge pipe D and from this into 
the intercooler E in which the air is cooled by water in the pipes F. 
The water is caused to circulate back and forth in this inter- 




Fig 



. 38. — Section of two-stage Ingersoll-Rand compressor. 



cooler. The air just compressed in cylinder G is forced through 
the intercooler into the cylinder H where it is compressed to a 
higher pressure. In this cylinder the inlet valves I are at the top 
of the cylinder while the discharge valves J are at the lower part 
of the cylinder head. Both sets of these are mushroom valves. 




VMS///* ■ , ,, ._, „ , , VS//W/A _ . V7///, 

Fig. 39. — Enlarged section of L. P. air cylinder of Ingersoll-Rand compressor. 

The air is finally sent to the air storage tank or aftercooler 
through the pipe K. 

Fig. 38, which is the Ingersoll-Rand class AA-2 compressor, 
illustrates one form of two-stage air compressor in which the 



AIR COMPRESSORS 



121 



driving steam cylinder is in line with the two air cylinders 
and the fly wheels are driven by outside connecting rods. The 
figure illustrates the method of bringing cool air to the compressor 
from a point outside of the engine room through the conduit L 
and at M and N are the water jackets to remove some of the heat 
of compression. 

Fig. 39 is introduced to show the valves of the low-pressure 
cylinder. In the periphery of each face of the piston are a series 

of slots distributed around the piston 
through which air can pass. Over this 
is fitted a ring which closes this open- 
ing and acts as a valve. This is known 
as the hurricane inlet valve. When 
the piston moves to the left at the be- 
ginning of a stroke the right-hand ring 
is moved from its openings and air can 
enter behind the right side, the com- 
pressed air on the left holding closed 
the valve on the left side. At the end 
of the stroke the inertia of the ring 
tends to close the right one as the pis- 
ton reverses and the left one will open 
as soon as the pressure of the air in 
the left-hand clearance space expands 
to atmospheric pressure. The dis- 
charge valves are ordinary mushroom 




Fig. 40. — Vertical after cooler and intercooler of Igersoll-Rand Co. 

valves with a light spring to close them and a tube on the 
back to act as a guide. 

Fig. 40 shows the construction of an intercooler, through which 
the air must pass from one stage to another and give up its heat. 
By the arrangement of baffle-plates and partitions the air and 
water are made to take a circuitous path so as to be more 
efficient in the removal of heat. The moisture which separates 



122 



HEAT ENGINEERING 



as the air is cooled is usually caught as shown in the figure so that 
it will not pass over into the next stage. 

In many cases an after cooler, Fig. 40, is used after the last 
stage to remove more of the moisture from the air and to cool it 
before it passes into the transmission line. In this way the air 
is of smaller volume and there is less friction. 

Fig. 41 illustrates the arrangement of the Taylor hydraulic 
air compressor. In this a system for the flow of water must 
exist. Suppose the dam A gives a head of H feet and this causes 
water to flow through the pipes B, C, D, E to the tail race F. 
The head and friction will cause a certain flow through the system 




Fig. 41. — Taylor hydraulic air compressor. 



and if the pipe is necked at C the velocity may become so high 
that a vacuum is formed at this point and air is drawn in through 
openings placed here. The air enters from G. This air mixes 
with the water and when the velocity is decreased in H the air 
separates out and rises to the surface of the water. This air is 
under pressure due to the head h on the chamber G. The air is 
taken out through the pipe I. 

The Humphrey apparatus is described in Engineering, 
Vol. LXXXVIII, p. 737, and in "Compressed Air Practice" 
by Richards. 

In all of these forms of apparatus the volume of air taken 
in might be the same while that discharged is determined by its 
pressure and temperature. To give some idea of the amount of 



AIR COMPRESSORS 123 

air used by tools or machines and the amount handled by the 
compressor it is customary to reduce the air to some one standard 
condition. The conditions taken are 14.7 lbs. absolute pres- 
sure and 60° F. temperature. This air is known as "free air." 
At times the temperature is assumed to be that of the atmosphere 
at the time of use, in which case the term free air refers to air at 
atmospheric pressure regardless of temperature. 

With this introduction the thermodynamics of compressed 
air will be considered. 

WORK OF COMPRESSION 

The amount of work required to compress Vf cu. ft. per minute 
of free air is shown by the diagram of Fig. 42 which assumes no 
clearance. The line ab is the atmospheric line and on account 
of the friction of the inlet pipe and P 
valves, the initial pressure pi is be- a 
low atmospheric pressure. The air 
is sucked in on the suction line cl 
and is compressed from 1 to 2 on 
the line pv n = const, and is then a 
driven out from 2 to d. The pres- c 
sure at 1 is pi and the volume is V\. 

The atmospheric pressure is p a and FlG - 42.— Indicator card from 

. ^ ^ compressor with no clearance, 

the free air Vf after throttling oc- 
cupies the volume Vi. The throttling action means constant 
heat content which for a perfect gas means isothermal action. 

Hence p a V f = Pl 7i (1) 

Some authors say that the air changes its temperature in 
entering the cylinder, but this cannot be appreciable as air is 
such a poor conductor of heat that it would take up little or no 
heat from the walls and the throttling of itself is isothermal. 

Now the work of compression is 

v^_v^ _ nV2 + piVi _ w (2) 

This quantity is negative since the work is done on the gas. A 
negative answer means that work is done on the gas. 
Reducing: 




124 HEAT ENGINEERING 

Now p 2 Y 2 = Pl^l" 

Hence work = ^-j Pl 7i [l - (^) V] (3) 

If desired p a V f may be substituted for piVi giving work 
to compress V f cu. ft. of free air from pi to p 2 as 

work = ^F,[l-g)^] (4) 

EFFECT OF CLEARANCE 

Now if there is clearance the indicator card takes the form 
shown in Fig. 43. The air remaining in the cylinder, d2', 

at the end of discharge expands 
from 2' to V on the return stroke 
preventing air from entering un- 
til V is reached. The amount 
of air then taken in is V'\ — 
V\, or V\ - Vi = Vi of the 
previous discussion. 

The net work required to 
drive the compressor is area 1" 
Fig. 43.-Effect of clearance. " 2"2'1' or area l"2"dc - 1'2'dc. 

This, assuming the same form 
of expansion and compression lines, gives 
Net work with clearance = 

-^i^'-^]['-©vl 




This is the same expression as (3) for the compression of Vi 
cu. ft. of air from pressure pi to p 2 , or in other words clearance 
has no effect on the work required to compress a definite volume 
of air. The effect of clearance is to decrease the amount of air 
taken in for a given displacement or to increase the displacement 



AIR COMPRESSORS 125 

for a given amount of air taken in. Thus in Fig. 42 the air taken 
in is Vi and the displacement D = Vi, but in Fig. 43 the air 
taken in is Vi but the displacement D = V\ + l'c — 2'd, a 
quantity greater than Vi. In other words the displacement has 
been increased. This increase of displacement means a larger 
cylinder in stroke or area of piston and hence there will be more 
friction and consequently more work will be required to drive 
the compressor, but the work indicated on the air for the com- 
pression of a certain amount is the same with or without 
clearance. 

The volume 2'd represents the volume of the clearance space 
and since this is usually expressed as a percentage of the dis- 
placement it may be represented as ID, where I is the percentage 
clearance and D is the displacement. 



Now l , c = 2'd( 2? )» 

w 

Hence Vi = D + ID - ID (^) » 



= D[l + i-z(g)»] (6) 

The term 1 + Z — Z I— J n is called the clearance factor. It 

may be represented by K h This term is only used to find the 
displacement if Vi is given or V\ if D is given. The clearance 
ratio I is known. 

EFFECT OF LEAKAGE 

If there is leakage around the piston, piston rod and through 
the valves the volume Vi required to deliver a given amount 
of free air Vf must be increased so as to care for this leakage. 
If the amount of free air delivered is expressed as a percentage of 
the free air taken in or as a ratio to the free air taken in, the per- 
centage is called the leakage factor/. Actual Vi = j ■• 

Although the leakage is effective during the compression, giv- 
ing less work as the compression is carried higher, it may be 
considered that this loss occurs at the upper pressure only and 



126 HEAT ENGINEERING 

Vi V f 

consequently increases the work by causing -j- or -/ to be 

substituted for Vi or V f . Thus 

(7) 

The effect of leakage is to increase the work. 

VOLUMETRIC EFFICIENCY 

The volumetric efficiency is defined as the ratio of the free air 
delivered to the displacement of the compressor. If the actual 
free air is used this is the actual volumetric efficiency while if 

the indicated free air is used, the indi- 
cated volumetric efficiency is obtained. 
Thus if ab is the atmospheric line from 
b an actual compressor, indicated volu- 
metric efficiency or apparent volumetric 

pressor showing volumetric e fficiencv = — = — • 
efficiency. ^ ' D ab 

AT VlVl 

Now xy = - 

Pa 

since the pressure is lowered by throttling action in which T 
is constant. The temperature of the cylinder walls may change 
the temperature of the air slightly but this is so slight that the 
air is considered to be at inlet temperature. Substituting for 
xy its value, the following is obtained 

Apparent vol. eff. = £g = g [l + I - l^\ (8) 

actual Vf f X ind. free air 




True vol. eff. 



D D 



= f X ind. vol. eff. = / g[l + I - I (g) ^] (9) 

True vol. eff. = — X leakage factor X clearance factor. (10) 



HORSE-POWER AND POWER OF MOTOR 

If Vi cu. ft. are required per minute the expression (7) gives 
the work per minute in foot-pounds. Consequently dividing 
the expression by 33,000 gives the horse-power shown by the air 



AIR COMPRESSORS 127 

card. This result must be divided by the efficiency of the air 
compressor, about 85 per cent, to 95 per cent, depending on size, 
before the horse-power to apply to the compressor is determined. 
To find the horse-power applied to the motor, be it a steam engine 
or an electric motor, the above result must again be divided by 
the efficiency of the motor. This may be about 90 per cent. 
Hence, horse-power applied to compressor forFi cu. ft. per minute 

___n piVi r 1 (PA—1 nn 

n- 1 eff. X/X 33000 L \pj "J K J 

indicated horse-power of engine drive 

_ji 1 

n — 1 eff. compressor X eff. of engine 

Pi7i 



['-©-] M) 



/ X 33000 

TEMPERATURE AT THE END OF COMPRESSION 

If the temperature is T x at the beginning of compression the 
temperature at the end of compression is 



(p 2 \2=± 



T 2 

\pi 

This is seen from the following: 

P2V2 71 = PiVx 71 
/BT 2 \ n (BTi\» 



p 2 l-nT 2 n = pJ-nTi" 

T,= T^y^r (13) 



COMPRESSION CURVE 



The compression curve desired for air compression depends 
on the way in which the air is to be used after compression. Air 
is such a poor conductor of heat, and at 60 or 100 revolutions 
per minute the action of the compressor is so rapid, that expan- 
sion in an engine or compression in the compressor practically 
takes place along an adiabatic 

pV lA = const. 

If air is compressed along an adiabatic 1-2, Fig. 45, and is 
expanded along an adiabatic in the engine this adiabatic will be 



128 HEAT ENGINEERING 

2-1, if there is no leakage nor cooling between the compressor 
and engine. If, however, the air is stored in a tank for some time 
before using in the engine the air at a high temperature T% is 
cooled to the original temperature TV This causes the volume 
to decrease so that the volume occupied by the air in the cylinder 
of the engine is V 2 f where 2' lies on the isothermal 12'. The 
expansion line in the engine is now 2'1' and the area 122'1' 
represents the loss of work due to the cooling in the tank. Al- 
though 1-2 is the best line for compression if the air is to be used 
before it can cool so that it will expand in the engine along 2-1 
giving no loss, it is evident that 1-2' would be the better line if 
the air is to be stored before using, since the temperature along 
this line is constant. Hence it is often stated that isothermal 
compression is the ideal and best method of compression. This 
is true, and true only, if storage of air 
is to be employed in the system, for 
other cases adiabatic compression may 
const, be the best method. If isothermal com- 
pression is used the work of compression 
becomes 12'dc, resulting in a saving in 

work of the area 122'. The loss when 
Fig. 45. — Saving due to . . . 

cooling on compression. the expansion in the air engine takes 

place is then 12T instead of 122'!'. The 

expression for the work with isothermal compression without 

clearance is 

Work = piFi log e ^ - p 2 V' 2 + PiVi 
but VzV'z = PiVi 




/. Work = p 1 V 1 log e ^ = - Vl V 1 log e £» (14) 

P2 Pi 

To approach this isothermal line in compression a water 
jacket is placed around the cylinder to remove heat, or water or 
oil is sprayed into the cylinder to reduce the heat. These methods 
are not very effective since n is changed only from n = 1.4 to 
n = 1.35. This saving is slight. The reason for this, as stated 
before, is the fact that the air is a poor conductor and also it is 
in contact with the cylinder walls a very short time. 

HEAT REMOVED BY JACKET 

The heat removed by the jacket is made up of two parts, that 
during the part of the stroke 1-2 and that during the part 2-d. 



AIR COMPRESSORS 129 

An expression may be written for the first part but no expression 
can be written for the part during the time that the piston 
moves from 2 to d. The temperature difference between the 
air and the jacket water is greatest dur- 
ing this time, but the cooling surface d 
in contact with the air is decreasing \^ pv \M m Con8fr . 

and this effect would tend to decrease the 
cooling effect while the greater tempera- c 

ture difference would increase the effect. 

T -, u ^ j , n ,i Fig. 46. — Card from com- 

If the effect is assumed to be the same pressor with jacket. 

as that during the portion of the stroke 

1-2, the total effect may be found by multiplying the effect 

Vi 1 

during 1-2 by -^ ^- or - 

u - n 




\P2/ 



x , , . J T 1 O ^2^2 - PlVl . V2V2 - PlVl 

IS ow heat removed on line 1-2 = j z -\ z 



[r^ + ^M'-©^] 



From the above and by considering the leakage factor /, the 
following is obtained: 
Heat removed by jacket 

1 r n-k i PiVi V, [VA 7 ^] (15) 
~ 1 _ (vA l -Ul-k){n- 1)J / L 1 \pj J 

SAVING DUE TO JACKET 

The work done by the compressor when the exponent is 
changed from k of the adiabatic to n is 

Work = _» Efxn _(a\==i] 

n — 1 / L \pi / J 

The work for adiabatic compression is 

The saving due to the jacket is the difference of these or 



130 



HEAT ENGINEERING 



WATER REQUIRED FOR JACKET 

If the heat removed by the jacket per minute is found, the 
water required is determined by assuming the possible range of 
temperature and then finding the water by 

~ heat per minute from jacket in B.t.u. 

Or = ', 1 

q -qi 
G = weight of water per minute. 
q' = heat of liquid at outlet. 
q'i = heat of liquid at inlet. 



(17) 



MULTISTAGING AND INTERMEDIATE PRESSURES 

Although jacketing is used the slight change in the exponent 
does not give a great saving in work and moreover for high pres- 
sures T 2 becomes so great even with a jacket 



t*-*©^] 



that the lubricating oil is apt to ignite and cause an explosion. 
To prevent this and to save work the air is compressed to a 




Fig. 47. — Two-stage compression. 

pressure less than that finally desired and discharged from the 
cylinder into a chamber containing a number of tubes carrying 
cold water. This chamber is called an intercooler and the tubes 
are made of such an area and arranged in such a manner that 
the air is brought to its original temperature at 1 before it is 



AIR COMPRESSORS 131 

sent to a second cylinder of proper volume in which it is com- 
pressed to its final pressure. This method of compressing in 
two cylinders of different sizes is known as two -stage compression. 
The action is shown in Fig. 47. V\ cu. ft. of air are drawn into 
the low-pressure cylinder and compressed to a pressure p' 2 . The 

— )~n~ is then discharged from this 

cylinder and through the intercooler until its temperature is 
reduced to 7\. The air is then drawn into the second cylinder 
which must be of such a volume as to take the air which occupies 
the volume V\ found on an isothermal through 1 at a pressure 
p' 2 . The air is then compressed to a pressure p 2 . The work in 
the two cylinders is given by 

Work in low-pressure cylinder = =- piT^i 1 — ( — J » 

Work in high-pressure cylinder = - p'zV'A 1 — (— r) n I 

Now p l V 1 = p\V\ 

Hence 

Total work =^^[2 -g)^ - ( J^ ] (18) 

The only variable in this expression is p' 2 . Hence to find the 
condition for minimum work, the first derivative with respect to 
p f 2 must be equated to zero. 

d . ■ . n r w-lV 2 \^l_il 

Wl (total work) = rT^i^H- -n7\TJ n n 

+ ^(^-^1 = 

n \p 2/ p 2 J 

., (E^)-il-(^)-^ 

\pi/ pi \p 2 / Pi* 

\p\P2l 

<n r 2 

^- = 1 
PlP2 

V'* = VPVP2 (19) 

That is, the work is a minimum if p\ is a mean proportional 
between p\ and p 2 . 



132 H^AT ENGINEERING 

Substituting this value of p' 2 in (13) the following results: 

to«. .- . ^ „v,[ 2 - pfl^-?. yy?| 

If there are three stages, proceeding in the same manner, the 
equation becomes 

T„„l »o,U . „-^W» "© -(^"-(ft)"] 

in which there are two variables, p' 2 and p'\. The conditions 
for a maximum or minimum are — — (total work) p » 2 = 

and — if- (total work) p > 2 = 0. These give p\ = y/pip' f 2 and 

p\ = VP2P2-, ^__^ = 

P*2 = \Zpi Vp f 2 P2 
PV = PlVp f 2P2 
pV = PlV2P2 
pV = P1 2 P2 



p\ = Vp?P2 == 3 (21) 

p" 2 = Vp&pfa = ^/piW = ^P^ 2 (22) 
For m stages there will be m - 1 intermediate variable pres- 
sures in the expression for total work and there will be m — 1 
partial derivatives to equate to zero giving, in the same manner 
as above, 

(23) 



p'« 


= 


7 y p m- 


>2 


or 


0"2 


m ^pl m - 


~ 2 P2 2 


or 









Pi w 



(24) 



p.'".-' = v^ 



i-i 



>ip 2 ' 

It is to be noted that the ratio of pressures on each stage is the 
same and each ratio of pressures on the various stages is equal to 

(— 2 jm. If these are substituted in the expression for total work 

the equation becomes 

Total work for m-stage compression = 

S ^,[l-©S?] (25) 



AIR COMPRESSORS 133 

This is the general expression for the work of an m stage com- 
pressor on the compression line pv n = const., for any value of n 
except 1. 

INTERMEDIATE TEMPERATURES 
Since the ratios of pressures on each stage are equal to (~) ™ 

it follows that the temperatures at the end of compression, T 2 , are 
all the same and equal to 

T 2 = 7\ (^- 2 ) ™ (26) 

For a single-stage compression between pi and p 2 

HEAT REMOVED BY INTERCOOLER 

In the intercooler the temperature is reduced at constant 
pressure from T 2 to 7\ and the heat is given by 

Heat from intercooler in foot-pounds = JMc p (T 2 — 7\) 

= j|^ c P (T 2 - ro 

If leakage is considered this becomes 

h 



Heat = — 



ifb -©"-] » 



AMOUNT OF WATER FOR INTERCOOLER 

In the above section the heat removed by the intercooler has 
been determined. The amount of water per minute required for 
the removal of this heat is found by assuming the temperature 
allowable at inlet and the temperature at outlet desirable and 
then computing by the formula, 

n heat per minute from intercooler in B.t.u. /ork . 

G = a' - a'- (29) 

q o q % 

G = weight of water per minute. 

q' = heat of liquid at outlet. 

q'i = heat of liquid at inlet. 



134 HEAT ENGINEERING 

The area of the surface of intercooler is found by methods of 
Chapter III. 

WATER REMOVED IN INTERCOOLER 

If air at pressure p a has a relative humidity p a and the weight 
of 1 cu. ft. of moisture to saturate the air at the temperature 2\ 
is m a , the total amount of moisture entering per minute is 

M w = PalTlaVf 

When this is sent to the intercooler and cooled to temperature 

u 
T\, after it is compressed to pressure p' 2 and volume Y\ or — j- Vf, 

V 2 

the amount of moisture held in the air, if saturated, will be 

V 2 

This is less than M w in most cases, so that the amount of 
moisture precipitated is 

M w - M' w 

If M' w is greater than M w the ratio -^r- will give the relative 

humidity of the air leaving the intercooler. This same method 
can be used to compute the moisture removed by the after cooler. 



EFFECT OF LEAKAGE IN MULTISTAGE COMPRESSIONS 

The leakage in a multistage compressor is a variable quantity, 
the leakage making the amount of air handled by the various 
stages different. Thus if 3 per cent, is the leakage in a single 
stage the leakage in a three-stage compressor might be 

1 - 0.97 X 0.97 X 0.97 = 0.088 = 0.09 approximately. 

That is, the amount of air to be handled by the lowest stage would 

Vi Vi 

be ttqTj ^ na t by the second stage tt-^t and that by the third would 

Vi 
be tt^j. Of course these differing amounts of air would change 

the theoretical discussion above but to a slight degree. Because 
of the similarity of relations for various stages the same pressure 
ratios will be used although the works on the various stages will 
not be the same. Hence in solving various problems the ex- 



AIR COMPRESSORS 135 

pression for total work will be used as derived before substituting 

y 

-j- for V h and using as / the mean value of the various /'s for the 

m stages. In computing the displacement of the various stages 
the correct value of / for each stage will be used. 



DISPLACEMENT OF CYLINDERS 

The displacement of each cylinder can be computed after the 
leakage factor and clearance factor are known for the cylinder. 

The leakage factor has been discussed in the previous section, 
and if 3 per cent, per cylinder is assumed the various leakage 
factors up to four stages may be taken as 88 per cent., 91 per 
cent., 94 per cent, and 97 per cent. The clearance factor is 
given by 



K : = 1 + I 
or = 1 + I - I 






If the clearance is the same on each stage the clearance factor 
will be the same for each stage. If, as is often the case, the value 
of I is small for the low-pressure cylinder and gradually increases, 
the value of Ki will gradually become larger for the high-pressure 

cylinders. Equations (6) and (30) show that as — or — becomes 

greater K t becomes less and so the effect of pressure on this factor 
is quite noticeable. With very high pressures on any stage the 
volume of expanded air at the end of the expansion part of the 
stroke is so great that only a small quantity of air will be drawn 
in. This is another reason for multistaging. 

Having K t and / for any stage the displacement is found by 

f x K lx p 2 x - 1 f x K lx v } 

In other words, the free air to be handled multiplied by the ratio 
of the pressure of the atmosphere to the initial pressure on the 
stage considered gives the volume of the free air when at the 
pressure of this stage and this divided by the volumetric efficiency 
for this stage, f x Ki X) gives the displacement per minute if Vf is the 
free air per minute. 



136 HEAT ENGINEERING 

If now the number of revolutions per minute is known and if 
the piston speed per minute allowable is assumed the length of 
stroke is known and then the area can be found after it is decided 
whether or not the compressor is double acting. 

Piston speed = 2LN = 200 to 700 ft. per minute. (32) 

D = (F h + F C )LN. (33) 

Fh = area of head end of air piston in square feet. 
F c = area of crank end of air piston in square feet. 
L = length of stroke in feet. 
N = number of revolutions per minute. 

SIZE OF INLET AND OUTLET PIPES AND VALVES 

The valve and pipe areas are such that the velocity of air is 
from 3000 to 6000 ft. per minute. Although these are high the 
loss in pressure is not excessive. The suction valve is open 
during a longer time than the discharge valve and for that reason 
it seems to be necessary to use larger areas on the discharge 
valves. On the other hand, the effect of the drop is more notice- 
able at the lower suction pressure and therefore the suction valve 
must have a large area. The valves are of about the same area. 
This area of each set may amount to 8 per cent, of the piston 
area for piston speeds of 300 ft. per minute while for speeds of 
700 ft. per minute 12 per cent, might be used. 

The pipes connecting cylinders or carrying air to or from the 
compressor should be designed from the allowable velocity if 
short, while for long pipes the drop in pressure, to be considered 
later, should determine the size. 

PERCENTAGE SAVING DUE TO MULTISTAGING OVER A SINGLE 

STAGE 

Work on single stage = — — r *-y-^ 1 — ( — ) ~^~ I 
Work on multistage = — — r -j~^\ 1 ~~ (— )~""»~ 

fl>-(-)"l 

/. per cent, saving = 100 - 100 ^ f\ „_ t (34) 

'•|>-(£)~] 



AIR COMPRESSORS 



137 



This saving is equal to the area li22'2i, Fig. 48, and of course 
it is the difference between the work of single-stage compres- 
sion and two-stage compression. The saving is equal to 



71 — 1 / L f m \pj fm \pj ^ J 



(35) 



UNAVOIDABLE LOSS OF COMPRESSION ON TWO STAGES 

The area 12ilil', Fig. 48, represents an area which cannot be 
regained by a single-stage engine and may therefore be called 
the unavoidable loss. It is equal to 12ied — Vlied. 




Fig. 48. — Saving due to multistaging. 

*~ - ;£W> -©*']- ^».-.[' - ©-] 



This is divided by / to give the total unavoidable loss with 
leakage. For three-stage compression the loss is really the work 



138 



HEAT ENGINEERING 



on two stages minus the work returned from one stage between 
pressures p" 2 and p±. 

This is shown in Fig. 49. 




Fig. 49. — Losses and gains for three-stage compression. 
WORK ON AIR ENGINE 

Air engines are usually of one stage and as the air expands its 
temperature falls so that at the end of expansion the moisture in 
the air freezes and in many cases clogs tip the exhaust. To 
prevent this heat may be applied to the exhaust pipe, multistag- 
ing may be used as shown in Fig. 51 or the air may be initially 
heated as shown in Fig. 50 by the dotted lines. This latter 
method produces such an increase in the work done that it will 
be carefully investigated later. 

The pressures between which the air engine operates are p' i 
and p\. These are different from p 2 and p± because there is a 
drop in pressure due to friction in the pipe line carrying air to 
the engine and in the valves entering the engine, thus changing 
Pi to p\. At exhaust the back pressure must be above the atmos- 
pheric pressure. 

If there is complete expansion in the engine and if the exhaust 
valve is so timed that the compression is complete, there is no 
effect of clearance on the work. Complete expansion or compres- 
sion means the carrying out of these actions until the final pres- 
sure is reached as shown in Fig. 52. The effect of clearance on 
the displacement is the same as that in the compressor and the 



AIR COMPRESSORS 



139 



same formula is used. In all theoretical discussions complete 
expansion is assumed and, for the present, the case of work 
or quantity of air for an engine with incomplete expansion and 
compression will not be discussed. 

The expression for work of the engine without clearance 
becomes, from Fig. 50, 




Fig. 50. — Single-stage engine card Fig. 51. — Two-stage engine 
wiih expansion line after preheating. card. 



1 -A; 



Work of single-stage engine = p'^V' \ + ^ x ' / __*"/ ' — pW'i 



In this case pW'i have been factored out because it is these 
two terms which are known in the case 
of the engine since the air is here at the 
original temperature T\; while at the 
lower pressure the temperature is low 
and would have to be computed by 



k 




before p\V\ could be found. 

(p\\ ^~L 
It is to be observed that ( ^Vj k is less than unity so that the 



Fig. 52. — Card with com- 
plete expansion and com- 
pression in engine. 



expression for work is positive, and moreover p\V\ — P\V\ 
or p a Vf so that either of these may be substituted, giving 

Work on single-stage engine = tt^ZT VaV A 1 — V" 7 /""*"" (^8) 



For a multistage engine the work becomes 

mk r /«'-v*-i- 



Wo* = 0- lP .V,[l-(^] 



(39) 



140 HEAT ENGINEERING 

If leakage is considered this reduces the work, giving 
mk _ r .. /©'A *zl 



r /M5T a [l ~gr)^] (40) 



The intermediate pressures are found as before 

LOSS DUE TO COOLING AFTER COMPRESSION 

The air is compressed to a temperature T 2 , for single or multiple 
staging, and on storing in the tank or pipe line its temperature 
is reduced to T\. The work which could have been done in one 
stage at the temperature T 2 would be given by 

Wak-^fMBT^l -($)*?] 

while after cooling to 7\ the work to be obtained is 

Wo* =^/^ [!-(£)¥] 

Hence the loss is 

Loss of work due to cooling 

-fV*[©--Jfc-$)T] <"> 

In this the first bracket is due to the compressor, while the 
second refers to the engine. 

UNAVOIDABLE LOSS DUE TO EXPANSION LINE 

Another unavoidable loss is due to the fact that the expansion 
line in the engine is of the form pV k = const, while that used on 
the compressor has been of the form pV n = const. This means 
that with a single expansion engine the area shown in Fig. 53 
by abc is unavoidably lost. This computation is made before 
cooling the air to the temperature Ti and while it is at the point 
P2V2. This loss is given by 



AIR COMPRESSORS 



141 



^ = W,-^!>-(£)-l-A!'-(S)-i] 



Since 



!T 2 T/ (p*\»=l T/ 




pvn= Const. 



(42) 



Fig. 53. — Unavoidable loss from difference in values of n in compressor and 

engine. 



LOSS OF PRESSURE IN PIPE LINE 

As the air flows through a pipe the pressure is decreased due 
to friction. There is only a slight change in velocity and hence 
this action is that of throttling in which the heat content and 
consequently the temperature of the perfect gas, air, remains 
constant. Now it is found that the drop in head when a fluid 
flows along a pipe line varies with the length of the pipe, with the 
square of the velocity and inversely with the diameter. This is 
usually expressed in feet of head of substance flowing. Since the 
velocity w of the air is only constant over a differential length of 
pipe due to the increase of volume as the air expands, the 
differential drop in pressure is all that can be expressed. 

— ah =Ts~at 
d 2g 

In this : b is a constant and equal to 

o.om + ^ + ^ + M^ 

w a dw 

w = velocity in feet per second. 
d = diameter of pipe in feet. 

1 Given by Green Economizer Co. 



0.02 approximately. (43) 



142 HEAT ENGINEERING 

g = 32.2. ft. per sec. per sec. 
dl = differential length in feet. 

dh = differential drop in head in feet of air. (dh is negative 
because it decreases as dl increases.) 

Now ^j, dh = dp 

Tjpfr — weight of 1 cu. ft. of air 

MBT 1 

and w = X ™ 

p r 

, F = area of pipe in square feet. 

M = weight of air per second in pounds. 

p b M 2 B 2 T 2 dl 
Hence — dp — 



or 



BT d p 2 F 2 2g 

J ^ = d^ -2g) dl 

(T is constant) 



pi 2 - p 2 2 _b^M^BT bM 2 BT M 2 L 

2 ~ ~d F 2 2g L ~ ^ L ~ d> (44) 

M 2 L 
p x *-p* = V'^ (45) 

Mean b" = 28 

In the equations (44) and (45), p x — original pressure in 
pounds per square foot, p 2 = final pressure in pounds per square 
foot, L = length in feet in which drop occurs. The formula may 
be used to find p 2 , M, or F, depending on what quantities are 
given. The formula 

Pi 2 ~ ?>2 2 = b" —^ 

can be changed to refer to the volume of free air at 60° F. instead 
of weight by dividing by the square of 

53.35 X 520 . . 
14.7 X 144 glVlng 

p L 2 -p 2 2 = b">^ (46) 



AIR COMPRESSORS 143 

Richards suggests &'" = iHzooo, if 

pi and p 2 = pressure in pounds per square inch. 
Vf = cubic feet of free air per minute. 
L = length of pipe in feet. 
D = diameter of pipe in inches. 

b'" = — — • 
2000 

The expression (44) may be simplified as follows: 

(Pi-P»)l— 2— y = d^ ~2g L 
V1 + V2 



2 
BT 

V 



= p = mean pressure 
mean volume of 1 lb. 



(Pi -P.) — = 5 — ^— ^ 

or Head drop in feet of air = ,~ m (47) 

where w m is the mean velocity since 

MBT TT _ 7 
-^- = Fand^ = w; 

Equation (47) is the formula used in hydraulics where the 
velocity is uniform. 

Equations (44), (45) or (46) may be used to find the length of 
pipe for a given drop and quantity by weight or volume for a 
given pipe, or D if the allowable drop in a certain length at a 
given discharge is known, or lastly the drop for a given discharge 
through a certain pipe of given length. The value of b is obtained 
by successive approximations in the formula (43) if not known 
from the given conditions. 

LOSS DUE TO LEAKAGE FROM TRANSMISSION LINE 

The leakage from the transmission line may amount to a 
large percentage of the air delivered if the line is not tight through- 
out its entire length. The leakage through a number of small 



144 HEAT ENGINEERING 

holes is larger than one would expect. The amount of leakage 
is found by Fliegner's formulae, depending on the pressure. 

M = 0.53 ^P= for p 2 < 0.5 Pl 
M = 1.060 F ^ jrtfr-P*) for p2 > 0>5 pi 
After M is found this may be reduced to free air by 

Vfl = ~1) 

Pa 

The loss is proportional to the quantity of air. 

Loss as per cent, of what should be obtained = 100 -^r * 

Vf 

LOSS DUE TO THROTTLING 

The friction action of the pipe line is the equivalent of throttling 
action and moreover in many cases air is stored in tanks under 
very high pressure to be used in engines or air motors at a re- 
duced pressure after passing through a reducing pressure valve. 
The action of this reducing pressure valve is throttling action 
and hence the temperature of the air remains constant during 
this action; j/zV'i is therefore the same as p\V\ or p a Vf. The 
pressure has been reduced. The available energy should have 
been 

but by throttling to p\ this is reduced to 



or 



Loss due to throttling = W - W" = 
k - 



-^>m"-m « 



This reduction has been due to the reduction of the upper 
pressure to a point near the lower pressure. 



AIR COMPRESSORS 



145 



GAIN FROM PREHEATING 

If air at a pressure p' ' % and temperature T\ is heated to the 
temperature T" 2 so that its volume is changed from Y\ to y" 2 , 
the heat added will be 

&JQ = JMc p (T" 2 - 7\) 



(49) 




Fig. 54. — Gain from preheating air. 
The increased work will be the area 2'2"1"1', or 
k 



&W = 



Eff.= 



AW k - 1 



p / 2(F " 2 _yg]|\_^Yj (50) 

this heat used in preheating is t 



k - 1 
The efficiency of this heat used in preheating is therefore 
k 



^^ p',(7". - V 2 ) 



= 1 



/ t \h-\ 



r\ - r\ 

If now this efficiency of preheating, 1 — \~r) k ' is higher 

than the overall efficiency of the system, i.e., the ratio of the work 
of the air motor to the work required for the compressor, it will pay 
thermodynamically to preheat. Of course the necessity for a 
warm exhaust or the possibility of increasing the output of a 

10 



146 HEAT ENGINEERING 

given quantity of air may make it advisable to use preheating, 
although the efficiency of this is not as great as the efficiency of 
the system. 

POWER AND DISPLACEMENT OF AIR ENGINE OR MOTOR 

In the case of complete expansion, the expression for work has 
been given. 

Work for single stage = ~-j p a fV f [ 1 - (^) V] (38) 

Work for single stage after preheating = 

Work for ra-stage expansion = 

A »*"/[> -(ft)-] <«) 

Since in these the volumes represent quantity per minute, the 
indicated horse-power may be found by dividing by 33,000 and 
this quantity multiplied by the mechanical efficiency of the motor 
or. engine will give the delivered horse-power. This efficiency 
may be taken at 75 to 90 per cent. 

If on the other hand the power required is known as well as 
the pressure available, the indicated power can be found from the 
assumed efficiency and then the work per minute. From the 
equations above, the volume of free air necessary for the engine 
or the amount of compressed air can be computed. 

The next step is to find the displacement of the engine. If 
V f is known, the amount of air per minute at the upper pressure 
is given by 

V't = H 1 (53) 

V 2 

If Y\ is known this can be used to find V\, the amount of air 
between the ends of the expansion and compression lines: 

F ' i = y ' 2 (S)^ (54) 

Displacement = -«- = — —, — r~ (55) 

* ['+<-«$)*] 

If there is leakage, D = -^- (56) 



AIR COMPRESSORS 



147 



Now all the above determinations are for complete expansion 
and compression. If, however, there is incomplete expansion or 
compression other calculations must be made. 

Let - be the proportional value cut-off and x be the value of 

compression expressed in terms of the displacement and assume 
that there is a clearance of ID. The work is found as the area 
of the card after the pressures at 3 and 6 (Fig. 56) are found. 




ID^ 




Fig. 55. — Diagram for air engine Fig. 56. — Diagram from air engine 
with complete expansion and com- with incomplete expansion and com- 
pression, pression. 



Pz = V 2 



r 



l + l 

l + x 



D.E 

eff. 



(57) 
(58) 



w . . + 33000 X D.H.P. 1 . 

Work per minute = : ~ = - Dp 2 + 

mech. " 



p 8 (l+Z)Z>-p' 2 (z +i)D 



1 -k 



= D -p' 2 -(l-x)p\ + 



- V \[l - x]D - 



p\(l + x)D - pelD 
1 -jfe 



P ,(l + I) ~ V^ (l + ^) - P'S +X)+ Pi 

1 -k 



(59) 



In this equation D is the unknown quantity and may be found 
for any given power required. The quantity in brackets is the 
mean effective pressure of the card. The amount of free air 



148 



HEAT ENGINEERING 



required for this motor is ascertained by determining the differ- 
ence in the weights of air at 2 and 5 and then reducing this to 
volume. If there is leakage, D is made D -£■ / in finding volume. 



V< = 



MBT, 

Va 


BT 1 

Va 


p' 2 (-+ i)d n , , n 

^ \r 1 ps(l + x)D 
BT 2 BT 6 




TiD 

Va 


T 2 T, 





(60) 



T 2 and T 5 are not definitely known on account of the action 
of the cylinder walls. On a test they could be determined, since 
the temperature at the beginning of compression is the same as 
that of the exhaust and from this the mass in the clearance space 
would be known. From the air supply the mass entering would 
be known and consequently the total mass at the point of cut- 
off or release. Since the pressures and volumes at these points 
are known, as well as the masses, the temperatures could be 
found. Although not strictly correct, T$ will be assumed the 
same as T 3 and T 2 will be assumed equal to TV This gives 



v,= 



D 



'4+') 






(61) 



With either of the cases above if D is known the stroke and 
diameter can be found as in the case of the compressor. 
Assume 2LN = 300 to 700 or 1000. 
Assume N and find L. 
Now D = (F h + F C )LN. 
From this F h and F c may be found. 

The output of the air motor is usually given in a problem and 
from this the i.h.p. may be found. 



h.p. output 
mech. eff. 



= i.h.p. 



(62) 



From this the size of the motor and the amount of free air may 
be found after the pressure limits and events of the stroke are 
assumed. After this is accomplished the drop in pressure in 
the supply line is found and finally the pressures, free air for the 



AIR COMPRESSORS 149 

compressor, size of compressor and the power to drive the same. 
The overall efficiency is found by 

~ n «. h.p. output of engine 

Overall eff . = t -. — , , , . (63) 

h.p. required to drive compressor 

This efficiency is found to be about 40 per cent, when worked out, 
although with leaks a lower efficiency is obtained. To show vari- 
ous values of efficiency a number of problems will be worked out 
later. 

FAN BLOWERS 

The turbo compressors and fan blowers not only give a com- 
pression of the air but in addition the velocity of the air at 
discharge is so great that there is an additional term for the gain 
of kinetic energy. In the turbo compressors the cooling is 
practically continuous, although it may be considered as an wa- 
stage compressor of a value of n of almost unity on account of 
the cooling effect of the metal and the water jacket. In this 
case the expression for work is 

-=^.[SI»-(D-!4 r a <«> 

For the fan blower the action is so rapid and the path so short 
that the action is assumed adiabatic and the expression is 

Wo^^{^[l-©^l + ^j (65) 
In this p 2 and pi differ by a small quantity. 

GOVERNING 

The fan and turbo compressor are of value because the quantity 
of air may be varied with the need, the pressure changing with 
the quantity and the power changing within known limits so that 
motors may be provided. With displacement compressors work- 
ing at a fixed pressure, the simple way of changing the quantity is to 
change the speed of the compressor. This may be accomplished 
with steam engine drives by a slight throttling of the steam. 
The point of cut-off is not changed and the pressure is only slightly 
changed because for a given delivery pressure for the air a definite 
steam pressure is required to give sufficient area on the steam 
card. The slight increase of pressure will overcome the friction 



150 



HEAT ENGINEERING 



and speed up the machine. When, however, the speed of the 
motor cannot be varied, as is the case with certain electric motors, 
the varying demand for air must be met in other ways. Among 
the ways suggested to care for a varying quantity of air the 
following may be mentioned : (a) delayed closing of inlet valve, 
(6) delayed closing of discharge valves, (c) throttling air on suc- 
tion and (d) changing clearance. These methods are all under 
the control of the governor. All except method (c) mean no 
change in efficiency. Fig. 57 shows methods (a), (b) and (c) 




\ Suction Valve Held Open 

\ for a Certain Time 

\ 




Clearance, 



Discharge Valve Held Open 
for a Certain Time 




Change in Clearance 

Fig. 57. — Methods of changing the discharge from an air compressor of 

fixed speed. 

as well as three cards for three different clearances. The clear- 
ance is varied by automatically connecting different chambers 
to the cylinder heads as the pressure rises. This is controlled by 
the governor as the pressure rises, necessitating a reduction in the 
quantity of air; the increase of clearance will cut down the 
quantity. When compressors are to be operated at different 
pressures, the work of the steam cylinder is controlled by a 
throttle governor, a variable cut-off governor or by a Meyer 
valve gear. 



AIR COMPRESSORS 
MOTORS 



151 



The engines using compressed air may be of the regular form 
of engine or its equivalent. Fig. 58 shows the section through a 
Little David riveter of the Ingersoll Rand Co. In this the 
throttle valve A is controlled by the handle B which is pressed 
by the thumb of the operator allowing air to enter. The valve C 




Fig. 58. — Little David riveter of Ingersoll-Rand Co. 

allows air to enter behind the piston D which is driven at a high 
speed against the shank E which drives the rivet. The valve C 
shown in black admits air to the groove at I and exhaust takes 
place through the passage running from the left end of the piston. 
After passing through passages in the valve the exhaust leaves 
through the outlet to the left of C. When G is uncovered air 




Fig. 59. — Ingersoll-Rand air drill (Little David). 

rushes through a small port and actuates the valve so that air is 
cut off from / and the port H is connected to outlet and the air 
to the right of the piston is discharged. A small port at this time 
discharges air into the large exhaust passage which is now cut off 
from the atmosphere and this forces the piston to the right. 



152 



HEAT ENGINEERING 



When the piston passes H the air remaining to the right and an 
amount which constantly leaks through a passage to the right of 
I is compressed, cushioning the piston. The increase of pressure 
here, and that due to a leak into the groove I after this is covered 
by the piston, causes the valve to reverse and the action to be 
repeated. 

Fig. 59 shows the construction of a drill which is the equivalent 
of four single-acting engines with rotary valves worked from the 
shaft while Fig. 60 shows a section through a rotary reamer or 
drill. This is a rotary engine. Air is admitted at A and causes 
the diaphragm or plate B to turn 
on the axis C. 

In Fig. 61 a preheater is shown. 
In this the passage of the air 
through the hollow portions of the 
heating surface raises the tempera- 
ture to a high point. The fuel 
may be coal, oil or gas. 

The arrangement and amount of 





Fig. 60. — Rotary air drill. 



Fig. 61. — Sullivan air preheater. 



surface in the intercooler is fixed by the principles of Chapter 
III, and will be discussed later in connection with a definite 
design. 

LOSSES IN TRANSMISSION 

The various losses discussed are shown in Fig. 62. 

abed = unavoidable loss due to two staging. This may 

be eliminated if a two-stage engine is used. 
efgd = loss due to leakage in compressor. 
gfh = loss due to change of line. 



AIR COMPRESSORS 

hfji = loss due to cooling. 
ijkl = loss due to leakage in line. 
knop — mnql = loss due to throttling. 

mrst = loss due to high back pressure. 



153 




Fig. 62. — Diagram showing various losses in air transmission. 

LOGARITHMIC DIAGRAMS 

Before solving problems it will be well to consider the con- 
struction of polytropics on logarithmic coordinates. If a curve 
of the form pV n = const, be plotted on the coordinates log p and 
log V it is found that the equation above takes the form 

log p + n log V = const. 

This is of the form x + ny = const., or the curve becomes a 
straight line and the value of n is the slope of the curve since 

dx 
dy 



— n 



If the coordinates of p and V of a curve are plotted logarithmic- 
ally with logarithmic coordinates and the points lie in a straight 
line, as in Fig. 63, the curve must be of the form pV n = const, 
and the slope of the line is the value of n. The value of n in 
the figure is to be 

n = 1.4 

In this figure the value of the logarithm extends from to 1 
and the numbers from 1 to 10, but the figure is constructed 



154 



HEAT ENGINEERING 



beyond that limit since the same variation in logarithm changes 
the numbers from 10 to 100, 100 to 1000, etc., or 1 to 0.1, 
0.1 to 0.01, etc. 



0.7 
7.0 
7H.0 
0.C 
CO 
00.0 
000 
0.5 
50 



0.4 
4.0 

LogP 4 ^ 



0.1 

10.0 °' 1 
L0 
10 



o 

s 


\ 


/ 


S 












































































s, 


















\ 


s t 


















\ 


\ 




















\ 






























\ 


\ 







0.4 0.6 0.0 0.7 0.8 0.9 1 

4.0 5.0 6.0 7.0 8.0 9.0 10 

40 60 60 70 80 90 100 





Log V 






\ Table 


from 




\ Log 


Diag. 




\ P = 10 


V=ioo 




\ P = 20 


V= 61 


p 


\ P = 40 


V= 37 




\ P = 60 


V" 28 




\ P = 100 


V= 19 




\ P = 200 


V= 13 




\ P = 250 


F= io 




V P = 300 


V= 9 




N. p B 600 


V = 5A 


1-40 


x. p « iooo 


V=3.7 




] 



1=20 



Fig. 63. — Logarithmic diagram with repeated coordinates for plotting 

pyi.4 _ con st. 



If in the figure from b at the end of ab, the vertical to c is drawn 
and then cd is drawn parallel to ab, and after taking d to e and 
drawing ef with one following fg, a series of lines may be put on a 
single figure which would require a much larger figure if a single 



AIR COMPRESSORS 



155 



line were drawn as in Fig. 64. Fig. 64 is less confusing than Fig. 
63, but it requires much more space. 

Such figures as the above are of value in constructing curves 
of expansion. If on a logarithmic diagram through the points 
log pi and log Vi, a line inclined at n = 1.35 be drawn, the values 







\ 


















x: 


















\ 


















v 


















\ 


















\ 




















\ 














































































































































































































_L\ 



0.1 0.2 0.4 0.6 0.8 1.0 2.0 4.0 6.0 8.0 10 

Log V 

Fig. 64. — Logarithmic diagram with continued coordinates. 

of p and V for any point may be found and the pV curve drawn 
with simplicity. 

To aid in computations of many of the above quantities charts 
and diagrams have been devised from which results may be 
obtained rapidly. A set of diagrams arranged by Professor C. R. 
Richards and Mr. J. A. Dent, based on temperature entropy 
logarithmic diagrams, is of great value. See Bulletin 63 of the 
Engineering Experiment Station, University of Illinois. 



PROBLEMS 

To apply the above formulae, a problem will be assumed as follows: 
Ten 5-h.p. air motors are to operate at 200 r.p.m. between 60.3 lbs. gauge 
pressure and 0.3 lb. gauge back pressure. Find the size of the motors if the 



156 



HEAT ENGINEERING 



piston speed is 200 ft. per minute and find the amount of free air required 
to drive these assuming a 3 per cent, leakage. Assume first that the com- 
pression and expansion are complete with a 5 per cent, clearance and sec- 
ond assume a cut-off at 0.35 stroke and compression at 0.1 stroke with 
5 per cent, clearance. Compression of this air is to take place in a two-stage 
air compressor from — 0.2 lb. gauge pressure to 135.3 lbs. gauge pressure. 
The compressor is 3000 ft. from the motors and the drop in the supply- 
main is to be 4 lbs. Find the size of the main. There are fifty H2-in. 
holes in the main ; find the leakage. The air temperature is 70° F. and the 
available water is at 60° F. The leakage in the compressors is 3 per cent. 
Find the air to be taken into the compressor to cover all leakages. Find 
the various efficiencies, losses and constants of the system. Would it be 
advisable to preheat the air to 350° F. ? In this problem the work will be 
done by the use of a 12-in. slide rule of Log. Log. form. 





Fig. 65. — Indicator card for air Fig. 66. — Indicator card for 

engine with complete expansion an engine with incomplete ex- 

and compression. pansion and compression. 

Problem 1. — Indicated horse-power of one motor. 



I.h.p. = 



h.p. 5 
eff. ~ 0.80 



= 6.25 



Friction 1.25 h.p. 

Problem 2. — Displacement and free air for two cases. 
(a) Complete expansion and compression. 



0.4 



6.25 X 33,000 = j^-j 14.7 X 144 X 0.97 7/ [ 1 - (j^f^|) "J (40) 
= 3.5 X 2118 X 0.977/ [ 1 - 0.631 1 



V/ = 78 cu. ft. per minute 
ri _l±7X78 ilM8 



75 
V'i = 15.26 X 
Ki=l + 0.05 - 0.05 X 



(53) 

©*-« ^ 

1_ 
(~ ) 1A = 1.05 - 0.158 = 0.892 (30) 

(56) 



D = ' q S q 2 = 52 cu. ft. per minute. 



AIR COMPRESSORS 157 



(6) Incomplete expansion and compression. 
1.4 



Pi = 75 Q Q5 j- i = 75 X ^ = 19.4 lbs. per square inch (57) 

[0 05 4- 11 *'^ 
' Q q 5 ' = 15 X 4.66 = 69.9 lbs. per square inch (58) 



(59) 



M.e.p. = 0.35 X 75 - 0.9 X 15 + 

19.4 X 1.05 - 75 X 0.4 - 15 X 0.15 + 69.9 X 0.05 
- 0.4 

= 26.25 - 13.5 -f- 21 = 33.75 lbs. per square inch 
6.25 X 33,000 = D X 144 X 33.75 

D = 42.5 cu. ft. per minute 
40 «i r /I 05\ 04 "I 

Vf = TiT^TiiL 144 x 75 x °' 4 " 15 x 144 Vox) x °' 15 J (61) 

= 2.89 30.0 - 3.3 = 77.2 cu. ft. per minute. 

If leakage is considered 

77.2 
V f = q-q^ = 79.5 cu. ft. 

The loss in power due to leakage is ( ttq^ — 6.25 ) =0.19 h.p. 

It is seen that although the displacement is larger for the complete expan- 
sion arrangement the amount of free air is slightly less. The difference 
between the two cases is not great. 
Problem 3. — Size of motors. 

(a) and (6). N = 200 
Assume 2LN = 200 

.'. L = ^ ft. = 6 in. 

(a) D = 52 = (A h + A c ) X 200 X H 

If the areas are equal 

a 52 u 

A =20^ sq.ft. 

52 X 144 _ . 
= 200 = 37 * 4 sq - m ' 
Assuming a 1-in. piston rod and tail rod 

A C yi. = 37.4 + 0.7854 = 38.2 sq. in. 
d = 6.95 in. = 7 in. 

Cylinder 7 in. X 6 in. 

(6) D = 42.5 cu. ft. = 2 X 200 X V2 X A 

42 5 
A = 200 x 144 = 30 ' 6 sc l- in - 
Acyi. = 30.6 + 0.7854 = 31.4 sq. in. 
d = 6.33 in. 

Cylinder 6.33 in. X 6 in. 



158 HEAT ENGINEERING 

Problem 4. — Leakage from pipe line. 

Pressure at one end = 135.3 + 14.7 = 150 

Assumed pressure for leakage = 150 

50 tv 150 

M = 0.53 XtX ^tt-t^ X 



4 32 X 32 V460+70 
= 0.132 lbs. per second 
60 X 0.132X53.3 5X530 iner „ . A 

Vfi = 14 7 y 144 = ^ cu " "■ P er mmute - 

Problem 5. — Total free air assuming case of complete expansion motors, 
(a) Air for motor = 78 cu. ft. per minute. 

(6) Leakage air = 105 cu. ft. per minute. 

(c) Total delivered free air = 885 cu. ft. per minute. 

(d) Total free air required for compressor = tt^-; = 942 cu. ft. per minute. 

(0.94 is leakage factor for two stages.) 
Problem 6. — Diameter of pipe line to carry air. 
_. 885 X 14.7 X 144 

M = 60 X 53.35 X 530 = U0 lbs " per secomL 
For first approximation b = 0.02 

/150- 146_ 2 X U42 = 0.02 X l.P X 53^5 X 530 x ^ ^ 

V ' 2 X 32.2 X ~ X rf 5 

16 

d 5 in inches = d h in feet X 12 5 = 1050 

d = 4.02 in.; use 4-in. pipe. 

b should then be checked 

53.35 X 530 



1.1 X 



148 X 144 



Hi X 0.7854 



= 16.7 ft. per second. 



i. nmo^ , ° 0274 , 0-00145 , 0. 012 „ oN 
b = 0.0124 + -M- + —^- + 16 7Q x yi2 (43) 

= 0.0124 + 0.0016 + 0.0044 + 0.0020 

= 0.0204 
The value of 0.02 is close. 
Using equation (46) 

150* - 146* - 8852 X ^ (46) 

150 14b - 200Q x D5 (4t>; 

D 5 = 990 

D = 3.97 in. or 4 in. 
The drop in pressure will not be 4 lbs. if a 4-in. pipe is used with b = 
0.0204 

2 0.0204 X l.l 2 X 53. 35X5 30 

(&) 

p 2 2 = 22,500 - 1238 = 21,262 
p 2 = 145.8 lbs. 
. * . Drop = 4.2 lbs. 
If a one-third increase in this is assumed a drop of 6 lbs. will be cared for, 



1442 X- : x2 ' //IV5 X3000 (44) 

2X32.2X^X 



AIR COMPRESSORS 159 

Problem 7. — Loss due to throttling from 150 lbs. to 144 lbs. abs. 

Loss - H 14.7 X 144 X 885 [ g) * - g) ^ (48) 

= 6,550,000[0.524 - 0.518] 

= 39,300 ft.-lbs. per min. = 1.19 h.p. 

Problem 8. — Loss due to throttling from 150 lbs. to 75 lbs. abs. 

0.4 



Loss = 6,550,000 

= 6,550,000 
= 6,550,000 



(g)"- 0.518] (48) 

0.631 - 0.518] 
0.113] 



= 740,000 ft.-lbs. per min. = 22.5 h.p. 

Loss if leakage occurs at higher pressure 

780 
= 22.5 X gg^ = 19.8 h.p. 

Problem 9. — Loss due to cooling from T 2 to 7\. 

0.35 

r^ 530 ® 1 ' 36 * 2 (26) 

1 

= 530(10.33) 7.7 

= 530 X 1.355 = 718° F. abs. = 258° F. 

0.35 0.4 

1.4 



Loss = Q 4 



X0.94X94 2X 14.7X144[ O * * « -l] [l- ( ») «] <«> 

= 6,550,000[1.355 - 1][1 - 0.518] 

= 1,122,000 ft.-lbs. per min. = 34.1 h.p. 



The use of values previously computed reduces the amount of computation. 
Problem 10. — Work on two-stage compression and intermediate pressure. 

0.35 

Work = 2 *}: S5 X 14.7 X 144 X *"'' 



i-mr] (2o) 



0.35 ~ ' ~ ~ 0.955 

= 15,120,000[1 - 1.355 

= - 5,370,000 ft.-lbs. per min. = 162.7 h.p. 
0.94 -j- 0.97 
0.955 = — '- k — ' — = leakage factor for work on two 

stages 

V't = Vl50 X 14.5 = 46.6 lbs. per sq. in. abs. (19) 

Problem 11. — Work on single stage. 

0.35 

*■* -S X 1A7X I* Xg[l -(&)**] W 

= 7,470,000[1 - 1.833] 

= 6,220,000 ft.-lbs. per min. = 188.5 h.p. 

Problem 12. — Saving due to double staging. 

Gain = 188.5 - 162.7 = 25.8 h.p. 



160 HEAT ENGINEERING 

Problem 13. — Saving due to jacket if on single stage (16). 

0.4 0.35 

n- U7V141Y 885fl.4r /150\IX| 1.35[ /ISoXOsII 

= 1,930,000{3.5[1 - 1.95] - 3.86[1 - 1.833]} 

= 1,930,000 { - 3.32 + 3.218} 

= - 197,000 ft.-lbs. per min. = 5.98 h.p. 

Problem 14. — Saving due to jacket if on two-stage compressor. 

Gain = 1,950,000J7[1 - (1.95) ]- 7.72[1 -(1.833)]} 
= 126,500 ft.-lbs. per min. =- 3.84 h.p. 

Problem 15. — Unavoidable loss in two staging (36). 

0.35 0.35 

Loss = ^ X 14.7 X 144 X ^[l - (j^j J |_1 - [ m ) J 

= - 7,560,000[0.355]Tl - f^\ 

= - 705,000 ft.-lbs. per min. = 21.25 h.p. 

Problem 16. — Loss due to change in expansion line of engine from n to k, 
starting from point at end of compression. 

- - »'■( At' - (S)""^]- M> - ©""HI m 



= 14.7 X 144 X 885 X 1.355 { 3.86 [l —3-^33 



- 3 ^ ~ ik 

= 2,540,000(1.755 - 1.706} 

= 124,500 ft.-lbs. per min. = 3.78 h.p. 

Problem 17. — Loss due to high back pressure in engine. 

= ^j X 780 X 14.7 X 144 [0.632 - 0.625] 
= 40,500 ft.-lbs. per min. = 1.23 h.p. 
Problem 18. — Loss due to leakage in compressor. 

Loss = available work without leakage — available 

work with leakage 
= [work of compression— unavoidable loss] X 

[1 — leakage factor] 
= [162.7 - 21.25][1 - 0.955] = 6.38 h.p. 



AIR COMPRESSORS 



161 



Problem 19. — Loss due to leakage in line. 

Loss = [work of compression— unavoidable loss— loss 

due to compressor leakage— loss due to change 

. , . , , v .leakage volume 
of exponent-^ loss due to cooling] , , , — -* 



= [162.7 - 21.25 - 6.38 - 3.78 - 34.1] 
= 11.5 h.p. 
Problem 20. — Power to drive compressor. 

_, ind. power 162.7 , ono . 

Power = r — «- = n nn = 180.8 h.p. 

mech. efT. 0.90 * 

Friction loss = 180.8 - 162.7 = 18.1 h.p. 

Problem 21. — Power of motor for compressor. 

180.8 



105 

885 



0.90 



= 200.1 h.p. 



Motor loss = 200.1 - 180.8 = 19.3 h.p. 
TABLE OF POWER 



H.p. 



Percentage 



Power supplied motors 

Loss motors 

Power supplied compressor 

Loss compressor 

Power supplied air 

Unavoidable loss 

Loss from leakage in compressor 

Loss due to change from n to k 

Loss due to cooling after delivery. . . . 

Available power supplied line 

Loss due to leakage in line 

Loss due to throttling to 75 lbs 

Available power at engine 

Loss due to high back pressure 

Loss due to leakage 3 per cent, of 65.1 

Available indicated work 

Amount loss in engine 20 per cent. . . 



21.25 
6.38 
3.78 

34.1 



200.1 
19.3 

180.8 
18.1 

162.7 



11.5 

19.8 



Amount delivered. 



Saving due to jacket 

Saving by double staging. 



1.23 
1.98 



97.2 



65.9 



62.70 
12.54 



50.20 



3.84 

25.8 



100.00 
9.6 



9.0 



10.6 
3.2 
1.9 

17.1 



5.7 
9.9 



0.6 
1.0 



6.3 
25.1 



100.00 

1.9 

12.9 



It is to be noted that there is 9.9 per cent, loss due to throttling and 5.7 
per cent, due to leakage. These might be saved, giving the overall effi- 
ciency about 40 per cent, in place of 25 per cent. The low efficiency has 
been due to the leakage from line and throttling. In good installations 
40 per cent, overall efficiency might be reached, 
n 



162 HEAT ENGINEERING 

Problem 22. — Heat removed by jacket (15). 
Heat = 



t _ @$4 -(KS) ] [ 941 x l4 - 7 * H 

0.35 

[-(s)1 

= - !- 73 X ^ X 13,830 X 144 X 0.355 

= - 436,000 ft.-lbs. per min. 
= — 561 B.t.u. per min. 



Problem 23. — Heat removed by intercooler (28). 



0.35 



=—-^®Xl47XUi)[l-(a^] 

= + 2,400,000 ft.-lbs. per min. 
= 3090 B.t.u. per min. 

Problem 24. — Water for jacket. 

Assume water rises from 75° to 90° F. in jacket. 

561 
G = Q0 __ 7 , = 37.4 lbs. per min. 

Problem 25. — Water for intercooler. 

Assume rise in temperature from 60° to 75° F. 

^ = 7 k _ fiQ = 206 lbs. per min. 

Problem 2$. — Mean At, assuming counter current flow from 60° to 75° 
F. for water and 70° to 258° F. for air, using Rensselaer formula. 

Ah = 258 - 75 = 183° F. 
Ah = 70 - 60 = 10° F. 

183 4- 10 
Note that this is quite different from „ = 97 ° F - 

Problem 27. — Surface required and arrangement. 
Assume water velocity = 3 ft. per sec. 
Assume air velocity = 10 ft. per sec. 

Value of K - 126 X 46 ' 6 X 144 X X v M~W\ S H _ 16 7 
Value ol & - iZb X 53. 35(164 + 460) X QQH X 6 - lb.7 

(44) Ch. I 
a . 3090X60 1flA .. 

Surface = ^y^T"^ = 169 sq.ft. 

Arrangement of M-in. tubes 12 ft. long with water inside. 

Area of 1 tube = ~ X | X 12 = 2.35 sq. ft. 



AIR COMPRESSORS 163 

169 

No. of tubes = 2~Q5 = 72 tubes. 

No. of tubes in one nest. 

Area of inside of tube = 7 X (t — jg) X t^ = 0.00258 sq. ft. 

Amount of water = 206 lbs. per min. 

Velocity of water = 3 ft. per sec, 180 ft. per min. 

No. tubes in parallel = 64 . 5 x 180 X 0.00258 = 7 tubes ' 

This means 10 nests, or the water will run from 7 tubes to 7 tubes 
so as to give a path of 120 ft. in cooler. 

14 7 614 

Volume of air per min. = (942 X 0.97) j^-g X ^0 = 340 cu. ft. 

340 
Area for air = 1 A . . an = 0.565 sq. ft. 
10 X 60 ^ 

Suppose the 7 tubes are made in a case so that the air may pass along. 
Area of passage 0.565 sq. ft. = 81 sq. in. To this the area of seven U-in. 
tubes or 3 sq. in. will be added giving 84 sq. in. for passage of air. 

Problem 28. — Size of compressors at 80 r.p.m. with 400 ft. piston speed. 

l 



Ki = 1 + 0.05 - 0.05 
= 0.932 



/ 150\ 277 
\14.5/ 



Low-pressure displacement = n 04 y a qqo = 1010 cu. ft. per min. 

1010 = 400 Amean 

Amt = 2.52 sq. ft. = 388 sq. in. 
Assume a 3-in piston rod; A = 7.06 sq. in. 
Agross = 363 sq. in. 
dcyi. = 21.5 in. 

i = 2fro = 2wft - . 

Size of compressor 21.5 in. X 30 in. 

High pressure displacement. 

Intermediate pressure = \/l50 X 14.5 = 46.6 lbs. 

00 r -I a n 

D = 0.97 X 0.932 X 46^6 = 308 CU " ft ' 
Stroke = 30 in. 

308 
Area net = -r^ = 0.77 sq. ft. = 111.0 sq. in. 

Area gross = 111.0 + 7.06 = 118.06 
d C yi. = 12.3 in. 
Size of high-pressure compressor 12.3 in. X 30 in. 
Problem 29. — The efficiency of preheating is 

0A 
) * A = 1 - 0.631 = 0.369 



(W 



With the leakage and throttling giving an efficiency of 25 per cent, on 
output or (25 + 6.3) = 31.3 per cent, on indicated work of engine it is 



164 HEAT ENGINEERING 

seen that this would pay. If there had been no throttling and the air 
were at a 150-lb. absolute pressure the efficiency would be 

OA 

Eff. = 1 - (j^j 1A = 1 - 0.518 = 0.482 

This, being better than 40 per cent., would pay. In most cases pre- 
heating will pay. It is well to note in passing that the efficiency of pre- 
heating depends on the range of pressure. 

TOPICS 

Topic 1. — What are the different methods of compressing air? Sketch 
machines for this purpose. Under what conditions is each used? 

Topic 2. — Explain the construction of a two-stage air compressor and give 
the reasons for the use of such apparatus. What is the purpose of the inter- 
cooler? Why is moisture apt to collect in the air space of the intercooler? 
How is it removed ? What is the peculiar form of inlet valve used on the low- 
pressure piston? 

Topic 3. — For what reason is air for a compressor taken from the atmos- 
phere and not from the engine room? Why are water-jackets used? Ex- 
plain the action of the Taylor hydraulic air compressor. 

Topic 4. — Derive the expression: 



wo*=^^[i-©^] 



Topic 5. — What is the effect of clearance? Prove this. Derive the ex- 
pression for the clearance factor. 

Topic 6. — Explain what is meant by volumetric efficiency. Derive the 
formula for true volumetric efficiency in terms of the leakage and clearance. 
Show how to find the horse-power to drive a compressor. 

Topic 6a. — Why is jacketing of value? Is this true under all conditions? 
Is jacketing very effective? Derive the expressions for the heat removed by 
the jacket, the saving by the jacket and the water required for the jacket. 

Topic 7. — What is multistaging ? Sketch a figure and show why this is of 
value. What is the function of the intercooler? How large should it be? 
Derive the expression for two-stage compression: 

w ork =^ P 4 2 -(f^-(^] 

Topic 8. — Using expression of Topic 7, show that the work is a minimum 
when 

p'l = VpiPi' ■ 

What are the conditions for minimum work for a three-stage compressor? 
Reduce the expression for work to a simple form. To what does the 
expression reduce for an m-stage compressor? 

Topic 9. — Derive the formula for the temperature at the end of compres- 
sion on the line pV n = const, when pi, p 2 , and T x are known. What is the 
value of the intermediate temperatures on a multistage compressor? Show 
that these are the same on each stage. 

Topic 10. — Derive the formula for the heat removed by the intercooler and 
that for the amount of water required? What is the effect of leakage in 
multistage compression? 



AIR COMPRESSORS 165 

Topic 11. — Explain method of finding displacement of the various cylin- 
ders of a multistage compressor to deliver a given amount of air. 

Topic 12. — Derive formulae for the moisture removed from the intercooler 
and for the condition of the air discharged from it. How are the inlet and 
outlet pipes and valves designed? 

Topic 13. — -Derive a formula for the saving due to multistaging. Derive 
the expression for the unavoidable loss due to two-stage compression. 

Topic 14. — Derive a formula for the work done in a single-stage engine. 
What is the temperature at the end of expansion? To what does the expres- 
sion for work reduce for a two-stage engine? 

Topic 15. — Derive an expression for the loss due to cooling after compression. 

Topic 16. — Derive a formula for the loss due to a difference in the lines of 
expansion in the engine and of compression in the compressor. 

Topic 17. — Derive the formula for the loss of pressure in a pipe line carry- 
ing air. Why is T assumed constant? What is the effect of leakage? 
How is the amount of leakage determined? 

Topic 18. — Derive the expression for the loss due to throttling and one for 
the gain from preheating. 

Topic 19. — Derive the expressions for the displacement of an air engine 
with complete expansion and compression and with incomplete expansion 
and compression to give a certain power. 

Topic 20. — Derive the expression for the work of a fan blower. Explain 
how the quantity of air from a constant-speed compressor is regulated. 

Topic 21. — Explain by means of a diagram the various losses which enter 
into an air-transmission system. Give the efficiency in terms of the ratio 
of two areas. 

Topic 22. — Explain the use of logarithmic diagrams in plotting curves of 
the form pV n = const. Explain the method of increasing the range of 
a diagram. 

Topic 23. — Sketch the forms of some of the air motors in common use and 
explain their action. 

PROBLEMS 

Problem 1. — Find the power to compress 1000 cu. ft. of free air per min- 
ute from 14.5 lbs. absolute to 60 lbs. gauge in a single-stage air compressor; 
n = 1.35, leakage = 3 per cent., clearance = 2 per cent. 

Problem 2. — Find the power to compress 1000 cu. ft. of free air per minute 
from —0.3 lbs. gauge pressure to 160 lbs. gauge pressure in a two-stage air 
compressor with n = 1.35, leakage 3 per cent, in each cylinder, clearance 3 
per cent. Find the intermediate pressure. 

Problem 3. — Find the size of the compressors in Problems 1 and 2 if they 
operate at 120 r.p.m. and are double-acting. 

Problem 4. — Find the temperature at the end of compression in Problems 
1 and 2 if the original temperature were 75° F. 

Problem 5. — Find the heat removed in the intercooler for compressor of 
Problem 2. Find the amount of water required if the water available is at 
60° F. and is arranged by counter current to leave at 80°F. Find AT 
assuming that the value of the coefficient of heat transmission is constant. 
Find the value of K by assuming the velocity of the air to be 30 ft. per second 
and the velocity of the water as 10 ft. per second and using Nicolson's 
formula. 



1 66 HE A T ENGINEERING 

Problem 6. — Using the variable parts of the expression for work find the 
relative work for single, two-stage and three-stage compression between 14.5 
lbs. and 290 lbs. absolute. Find the clearance factor for these assuming 3 
per cent, clearance in each cylinder. Find the temperature at the end of 
compression in each case assuming 75° F. as the original temperature. 

Problem 7. — Find the size air main to deliver 3000 cu. ft. of free air per 
minute when compressed to 145 lbs. absolute, if the temperature is 75° F. 
and the drop in pressure allowable is 2 lbs. in 160 ft. 

Problem 8. — Find the efficiency of transmission if 1000 cu. ft. of free air 
per minute is compressed to 190 lbs. gauge pressure, transmitted 800 ft. 
in a tight main and used after storage in engines at 150 lbs. pressure in a 
single-stage engine. Assume all constants needed. Is the result the 
probable efficiency or the maximum efficiency? 

Problem 9. — Find the various losses in the system described in Problem 8. 

Problem 10. — Three U-in. holes are in a pipe line carrying 1500 cu. ft. of 
free air per minute compressed to 125 lbs. gauge pressure and at 80° F. What 
is the percentage loss due to these holes? 

Problem 11.— The wet bulb reads 70° F. in 80° F. weather when the ba- 
rometer stands at 29.85 in. This air is compressed to 50 lbs. gauge pressure 
in the intercooler at 80° F. Has any moisture been precipitated? If not 
what is the relative humidity of the air? If precipitation occurs how much 
precipitation occurs per minute if 1000 cu. ft. of free air are handled per 
minute? 

Problem 12. — Find the size of an air engine to develop 15 h.p. (a) with 
complete expansion and compression, and (6) with cut-off at 30 per cent., 
compression 20 per cent., clearance 5 per cent, pi = 125 lbs. absolute, pb = 
14.8 lbs. absolute. R.p.m. = 125. 

Problem 13. — Air at 1500 lbs. absolute pressure is stored in a tank and when 
used in an engine it is throttled to 150 lbs. absolute pressure. The tempera- 
ture in the tank is 80° F. What is the temperature at the end of complete 
expansion in the engine to 15 lbs. absolute pressure ? How much of the energy 
in the tank is available in the engine after throttling? 

Problem 14. — In Problem 13 the air on leaving the throttle valve is heated 
by a vapor lamp to 450° F. By what per cent, is the power of the engine 
increased by this? What is the temperature of discharge from the engine? 

Problem 15. — An air compressor of Problem 1 has the compression line 
reduced to pv U35 = constant by the jacket. How much heat does the jacket 
remove? How much water does this require if the water range is from 70° F. 
to90°F.? 

Problem 16. — One thousand cubic feet of free air per minute is compressed 
from 15 lbs. pressure to 150 lbs. pressure (absolute) on two stages. The 
original temperature was 60° F. Find the final temperature of discharge. 
Find the percentage loss if this air is allowed to cool to 60° F. before it. is 
used in the air motors; n = 1.35. 

Problem 17. — How much loss occurs in Problem 15 due to the fact that the 
expansion in the engine is on the line pv k = const, instead of pV 1 - 35 = 
const. ? Express this as a percentage. 

Problem 18. — Find the values of p and v at several points on the line, 
p V 1 - 25 = const., passing through the point p = 165 lbs. V = 3 cu. ft. by 
means of a logarithmic diagram. Use pressures of 150, 100, 50 and 25 lbs. 



CHAPTER V 
THE STEAM ENGINE 

On account of the numerous improvements made upon it 
during the two centuries of its use, the steam engine was, until 
quite recently, the most important heat engine using steam. 
The steam turbine has taken this place at present due to the 
decreased cost of production, the greater concentration of power, 
and the good efficiencies over a wider range of load. 

The engine was developed without any theory of thermodynam- 
ics by Worcester, Papin, Savery, Newcomen and Watt, and to 
account for the action of the engine in later times, Rankine in 
England and Clausius in Germany developed a mathematical 
theory of heat. The later development of the engine is connected 
with the names of Corliss, Porter, Reynolds, Willans, Sultzer 
and Stumpf. The engine has been improved by increasing the 
Carnot efficiency through the use of higher initial pressures and 
superheat and by lowering the back pressure, while the use of 
jackets, reheaters, superheated steam, four valves, heavy lagging, 
multiple staging and flow in one direction have all had their 
effects on its practical efficiency. 

In the steam engine (Fig. 67) steam containing heat is ad- 
mitted to the cylinder A by the valve B, so that it moves 
the piston. The valve B is operated by the valve rod C 
from an eccentric or crank D so that when the piston E has 
reached a certain point steam is cut off from the cylinder and 
the steam is allowed to expand to a point near or at the 
end of the stroke (release point) when the valve is so moved 
as to allow the steam to escape to the outside. The steam 
admitted to the other end or the inertia of the fly wheel causes the 
piston to return in its stroke forcing the steam out of the cylinder 
until a point near the end of the stroke (point of compression) is 
reached when the exhaust is closed and the movement of the 
piston compresses the steam until the valve again connects this 
side of the cylinder to the steam chest F and steam pipe G so 
that steam enters once more. In theory the indicator card 

167 



168 



HEAT ENGINEERING 




THE STEAM ENGINE 



169 




T3 

a 
o 

I 

O 

o 

'I 

a 

02 



170 



HEAT ENGINEERING 



will appear as in Fig. 68. The expansion and compression are 
complete. The point of cut-off is at 1, release at 2, compres- 
sion at 3 and admission at 4. The quality of the steam at 3 
is the same as that at 2, hence the condition at 4 is the same 
as that at 1 if the lines 1-2 and 3-4 are assumed to be adiabatics. 
3-4 is the compression line of the steam in the clearance space 
4-5. If these are adiabatics with M pounds of steam on 3-4 
and M + M pounds of steam on 1-2, Fig. 69, with no clearance 
space and with the weight of steam M on the line 1-2 between 
the same pressures and qualities at the limiting points will 
have the same area and represent the cycle with no clear- 
ance. This is true because in theory the clearance steam ex- 





Fig. 68. — Indicator card from en- 
gine with complete expansion and com- 
plete compression. 



Fig. 69. — Indicator card with 
no clearance. 



pands and contracts along the same line requiring and giving no 
work. M is called the clearance steam and M is called the 
working steam, steam supply or boiler steam. 



CYCLE OF STEAM ENGINE 

Fig. 69 then represents the theoretical form of cycle used on 
the steam engine. It is known as the Rankine or Clausius 
cycle although the former name is employed at times for the 
cycle shown in Fig. 71. The difference between the Carnot 
cycle and the Rankine cycle is that in the former cycle the 
working substance is supposed to be in the cylinder and heat 
only is added, while in the latter substance is added and with it 
heat. On the Carnot cycle, if represented by Fig. 68, the quality 
changes from that at 4 to that at 1, growing greater, while it 
decreases from 2 to 3, hence 1-2 and 4-3 are not similar adia- 
batics. For the Rankine cycle the quality on 4-1 is constant, 
the variable being the mass and this is true on 2-3. In both 



THE STEAM ENGINE 171 

cases 4-1 and 2-3 are isothermals if saturated steam is used. 
The heat added on the Carnot cycle is Mrdx and on the other 
is xrdm, the total amount being the same in each. 

In Fig. 69 the exhaust steam could be used to heat an equal 
weight of water to the temperature of exhaust for boiler feed, 
since this could be done by the heat of the liquid of the steam 
even if the water supply were at 32° F. With the heat of vapori- 
zation in addition for the vapor part of the mixture, this and 
even more could be done. This additional amount is of no value 
for the operation of the engine although for warming other 
water it has a great value. The point which must be grasped 
by the student is the fact that the heat in the exhaust could, if 
properly used, heat the feed necessary for the working steam to 
a temperature of the exhaust pressure. For this reason, the 
datum plane from which heat is to be measured is always taken 
as at the temperature of the exhaust. 

HEAT AND EFFICIENCY OF CYCLE 

The heats added on the different lines are: 

M(i! - q' ) on 4-1 (1) 

on 1-2 
M(i 2 - q'o) on 2-3 (2) 

on 3-4, since M = 

(Note. — q' is used rather than q f 2 because is used to refer 
to the datum in all cases.) 

.'.AW = Q 1 -Q 2 = MKt! - q' ) - (i 2 -q' )] = ilffa - i 2 )(3) 

It is to be remembered that i\ and i 2 are the heat contents at 
two points on the same adiabatic and hence they are found in 
the same entropy column or line. 

Q 1 = M[n - q' ] 

Theoretical eff. = r? 3 = r (5) 

ii- q w 

This is sometimes called the Rankine efficiency. 

In some texts the work of pumping the water into the boiler 
from 4 to 1 is considered giving net work 

i'i — i 2 — A(p! - p 2 )v' 



172 



HEAT ENGINEERING 



The last term is small. It really does not belong to the engine 
being one of the charges against the boiler. 

7\- T 



Now, the Carnot eff . = 771 = 



m 



T, 

m 



If M pounds of steam are required per horse-power hour, the 
work = 2546 B.t.u. and the heat is M(i x - q' ). 

2546 



lb 





(6) 



Fig. 70.— T-S diagram of 
Rankine cycle. 



Fig. 71. — Rankine cycle with incom- 
plete expansion. 



The T-S diagram for this is an aid in studying the cycle. 
It is shown in Fig. 70. In this, points 3 and 4 come together. 
The heat added per pound from 3 to 1 is 
6-4-3-1-5 = ii - q' 

Heat 12 = 

Heat 23 = 6-3-2-5 = i % - q'o 

Work = 64315-6325 = 3412 = i x - i 2 
4-a-l-2 would show the Carnot cycle. 

If the expansion is not carried to the back pressure line it is 
called incomplete expansion and the cycle using such is then 
sometimes spoken of as the Rankine cycle. The best way to 
study this cycle is by dividing it into two parts by the line 2-6, 
Fig. 71. 

Area 5126 = M{i x - i 2 ) 

Area 2346 = M[A(p 2 - Ps)v 2 ] 

AW = 451234 = M[(n - i 2 ) + A(p 2 - p 3 )v 2 ] (7) 



THE STEAM ENGINE 173 

ii - i 2 + A (p 2 - ps)v2 , Q . 

773 = u - tf' " (8) 

7\- T 



11 




^73 


T, 


2 




771 










2546 



775 - Af(*x - q'o) 

773 for the engine cycle is sometimes called the Rankine efficiency 
of the cycle or this term is applied to equation (5). 

In the above equations, i\ and i 2 are heat contents on the 
same adiabatic, hence, if the entropy is known for point 1, i 2 is 
found at pressure 2 in same entropy column as i\ if there is a 
table or on the same entropy line from a chart. If there is 
neither table nor chart which can be used, the formula for the 
adiabatic is to be used to find the quality at 2 and from this, i 2 . 

Vi = Mx x v\ (To find Zi). 

s\ + ~y = s' 2 + -|r (To find x 2 ) 

V 2 = Mx 2 v"* (To find V 2 ). 

ii = q'i + XiTx (To find n). 

ii = q'2 + x 2 r 2 (To find i 2 ). 

All necessary terms are known for any of these formulae if 
taken in the order given. The value of Xi being theoretical and 
therefore the highest possible, is that of the boiler steam and 
should be so used. 

STEAM CONSUMPTION 

From equations (3) and (7) the amount of steam per horse- 
power hour theoretically required is given by 

M=jm (9 ) 

tl - 12 
M = 7 , . , r— jr (10) 

^l - ^2 + A (p 2 - Ps) v" 2 

To illustrate the applications of the formulae, suppose an 
engine uses 26 lbs. of steam per horse-power hour at 110 lbs. 
gauge pressure with x = 0.98 and has a pressure at the end 
of expansion of 15 lbs. gauge and a back pressure of 0.5 lbs. 
gauge. It is desired to find the various efficiencies. 



174 HEAT ENGINEERING 



rom Peabody's Temperature 


From Marks and Davis : 


Entropy Tables 


Charts 


7)1= 110 + 14.7 = 124.7 


Vi = 124.7 


h = 344.2 


si = 1.563 


8i = 1.563 


fi = 1173 


it = 1172.5 


p' 2 = 29.7 


p 2 = 15 + 14.7 = 29.7 


s 2 = 1.563 


s 2 = 1.563 


f 2 = 1065 


fa = 1065.8 


x 2 = 0.897 


v 2 = 12.55 


z; 2 = 12.5 


q' of (0.5 + 14.7) lbs. = 181.9 





t = 213.7 

By computation the following results, using the tables of 
Marks and Davis: 

Vi = 129.7 

it = 318.4 + 0.98 X 872.7 = 1173.4 

*i = 0.4996 + 0.98 X 1.0820 = 1.56 

p 2 = 29.7 

s 2 = 1.56 = 0.3673 + x X 1.3326 

1.1927 
X2 = 0326 = °- 895 
it = 218.2 + 0.895 X 945.6 = 1065 
v 2 = 0.895 X 13.89 = 12.42 
q' = q f of (14.7 + 0.5) = 181.6 

Any of the three sets of values could be used. Using the 
first 

344.2-213.7 130.5 
771 = 344.2 + 459.6 = 803^ = 16 ' 2 per Cent * 

1172.5 - 1065.8 + 4^(29.7 - 15 - 2 ) 144 X 12 - 55 

11 o 

713 = 1172.5 - 181.9 

1399 no 
= 9906 = 142perCent - 

14.2 
= y^-~ = 87.2 per cent. 



V2 



2546 ?546 

775 " 26(1172.5-181.9) " 2580 ~ ^ per Cent ' 

9.9 
174 = 1T2 = ^ per cen ^ 



THE STEAM ENGINE 



175 



The results of the above computations show that the efficiency 
of the theoretical cycle is more than 87 per cent, of that of the 
most perfect cycle but that the actual engine only utilizes 70 
per cent, of the amount which should be available theoretically. 
For these reasons there is not much chance of improving the 
cycle of the steam engine but to make 774, the practical efficiency, 
greater the cylinder is covered with some non-conducting material, 
a steam jacket is used or superheated steam may be employed. 
These all tend to cut down the transfer of heat to and from the 
cylinder walls and thus cut down the amount of steam required. 

TEMPERATURE— ENTROPY DIAGRAMS 

The cycle is shown on the T-s diagram, Fig. 72. The 
free expansion from 2 to 3 cuts off the corner of the figure as 
shown by 2-3. This figure and Fig. 70 are used to represent 





Fig. 72. — T-S diagram of Rankine 
cycle with incomplete expansion. 



Fig. 73. — Effect of increasing tem- 
perature range. 



the Rankine cycle without and with complete expansion. They 
are not true representations since the lines of Figs. 70 and 72 
represent additions of heat and not of substance as is the case 
on the actual cycle. Because the heat added in the figures on 
the various lines represents the heat added on the lines with the 
substance the figures are used to represent this cycle. More- 
over, the area of the figure 1234 of Fig. 70 or 123456 of Fig. 72 
represents the difference between Q\ and Q2 and hence repre- 
sents the work. 

The efficiency is represented by the area of the cycle divided 
by the total area beneath 4-1 in Fig. 70 or 72 and hence, if the 
temperature is increased, both of these areas become greater and 
the efficiency is increased. This is shown by 61-I" of Fig. 73. 

46 1 1"234 461234 
m ' ~ 846x1 "7 > 84617 



176 HEAT ENGINEERING 

If, now, the lower temperature is lowered from 3 to 3', the 
denominator of the fraction is changed a slight amount, 8'4'48, 
while the numerator is changed by the area 433 / 4 / . In this way the 
efficiency is increased in most cases. If the free expansion is so 
great, as shown in Fig. 73, that the distance 3-4 is small, it maybe 
that the added area, 433 '4', may not be greater than 8'4'48 
and the efficiency is not increased by lowering the back pressure. 
As the free expansion 2-3 becomes less, as 2'-3" (which means 
that the expansion 1-2 is more nearly complete), the line 3"-4 
is so long that the gain in efficiency due to a decrease in back 
pressure is great. The increase in pressure to give a perceptible 
increase in temperature for the boiler steam is so great that the 

pressure must be increased considerably 
to raise the line b-1 an appreciable 
amount. Hence the gain in efficiency 
from an increase of pressure is not as 
marked as a slight decrease in the lower 
pressure of the cycle when the tempera- 
ture change is more rapid. Thus, to de- 
crease the back pressure from 15 lbs. to 

2 lbs. abs. means a change of 87° F. 

9 8 s while an increase from 100 lbs. to 113 

Fig. 74—T-S diagram ^s. means an increase of 10° F., from 
ot Rankme cycle with • „ _ „ 

superheated steam. 150 lbs. to 163 lbs., 8 F., while from 200 

lbs. to 213 lbs. means an increase of less 

than 6° F. It must also be remembered that a decrease in the 

back pressure with much free expansion does not necessarily 

give an increase of efficiency. 

In order to increase the efficiency by increasing the upper 

temperature without an increase in pressure superheated steam 

is used. Theoretically it will be seen that although this does 

increase the efficiency there is not much gain because this heat 

is not added at constant temperature. Fig. 74 illustrates 

this case 

Q 1 = <)cbl8 = q\ - q'o + r + fc p dt = i x - q' 

Q 2 = 94328 = i 2 - q' - A(p 2 - p 3 )v 2 

Q x - Q 2 = work = 45c6 1234 

Now the triangle 1-d-b is the part of the figure which repre- 
sents the increase due to superheat and being added to5 cbd 234: 



THE STEAM ENGINE 



177 



as well as to 9 c b d 8 it increases the efficiency, having the greater 
effect on the small numerator. This reasoning is made evident 



by observing that the efficiency 



5 cbd 234 . 



is about the same 



95 cbd 8 

as the efficiency of the cycle of Fig. 71 with saturated steam and 
the same pressure ranges and the expression for the latter is 
made into that for Fig. 74 by adding b-l-d to numerator and 
denominator. 



EFFECTS OF CHANGES 



These facts have been shown by the late Prof. H. W. Spangler 
in his Applied Thermodynamics. The curves of Figs. 75, 76 
and 77 show how the theoretical efficiency is increased as the 
pressures and superheats are changed. 



40 i 

3056 

S 20tf 

e 

H 10* 



Nou-Coudeusing 



40 ^< 
30jS 

lOjf 
















s 


J 



















40 f 
S 30$ 












s 

• 3 20 i 



0) 

1 10 # 

A 
H 





























100* 125^ 150^ 175# 
Gage Pressure 



74" 15" 221/2" 30 c 
Inches Vacuum 



0° 50° 100° 150° 200 
Degrees of Superheat 



Fig. 75. Fig. 76. Fig. 77. 

Fig. 75. — Effect of change of steam pressure on efficiency, complete ex- 
pansion and saturated steam. 

Fig. 76. — Effect of change of back pressure on efficiency (150 lb. steam 
1° superheat), complete expansion. 

Fig. 77. — Effect of change of superheat on efficiency (151 lb. steam to 
28.5 in vac), complete expansion. 



These figures are all drawn for complete expansion and, al- 
though in all cases the effect of vacuum increase is greater than 
the effect of the same change in the initial pressure, the effect 
is not so marked as that shown in Fig. 76 when there is incom- 
plete expansion. These figures are all drawn from theoretical 
considerations. 

Although the effect of superheat is theoretically slight, its 
effect on the actual efficiency may be much more pronounced 
since it cuts down certain losses which occur in the cylinder and 
therefore its practical effect is very valuable. 
12 



178 HEAT ENGINEERING 



TESTS OF ENGINES 



To actually determine the steam consumption of an engine a 
test is made while the load is kept constant on the engine. If 
possible the exhaust steam is condensed in a surface condenser 
and weighed at regular intervals and at the same time indicator 
cards are taken and readings are made on the scales of the 
Prony brake, or watt meter if the engine is used to drive a genera- 
tor, on the calorimeter to determine the quality of the steam, 
on the pressure gauge and barometer, on the temperature and 
pressure of the exhaust steam, on the temperature of the con- 
densate and on the revolution counter. From this data, as will 
be shown, the steam per horse-power hour, the pressure and 
quality of the steam supply, and the pressure of the exhaust 
may be found to be used for the actual thermal efficiency. 

CALORIMETERS 

The calorimeter used to determine the quality of the steam 
may be of three forms: the electric, the separating and the 
throttling calorimeters. The latter is shown in Fig. 78. The form 
here shown is the Barrus Universal Throttling Calorimeter, so 
called because by the addition of the drip pot C it may care for 
steam of any moisture content. In this a sample of steam is 
taken preferably from a vertical steam pipe A. In horizontal 
pipes it has been found that much of the moisture separates out 
and flows along the bottom of the pipe so that a fair sample 
cannot be obtained. The sampling tube B is a pipe closed at 
the end and containing a number of small holes around its circum- 
ference and along its length to take steam from various parts of 
the pipe. This steam is carried into the drip pot C, which is in 
reality a separator, where most of the moisture is taken from 
the steam. The height of the water in the drip pot is shown 
by the glass gauge D and the pressure is shown by the gauge E. 
Steam then passes over through a pipe and around a thermometer 
well containing the upper thermometer F and from here it 
passes through a small hole in the throttle plate G to the ther- 
mometer well containing the lower thermometer H and from here 
to the atmosphere. The mercury gauge I shows the pressure 
around the lower thermometer well. The throttle plate is held 
between flanges but insulated from them by some non-conducting 
material. 



THE STEAM ENGINE 



179 



To find the amount of water collected in the drip pot a string 
is tied around the glass gauge and when the water reaches that 
level the time is noted. The water in the drip pot is drawn off 
at any convenient time into a cup of cold water J so that the 
water level falls below the string. This is done by opening the 
valve K and putting the end of the drip pipe beneath the water 
level in the cup so that none of the hot water from the drip pot 




P 



H 







ntnStoi 




Fig. 78. — Barrus universal calorimeter. 



can evaporate on reaching a place of atmospheric pressure. 
The cup is lowered after closing K and after catching all the 
drip its change of weight will give the weight of water drawn 
off. The time at which the water again reaches the level of the 
string is noted and the interval beween the two times of passing 
the mark will give the time taken for this water to be caught. 
This is reduced to pounds per minute or second. Call this 
m pounds. 



180 HEAT ENGINEERING 

To find the weight of steam passed through the throttle hole 
in the same interval of time, this steam is exhausted into water 
and condensed or, what is quite customary, the hole is measured 
and Napier's formula 

70 
is used. 

If the separator were perfect the quality of the steam coming 
from the pipe would be, 

M 



M + m 



(11) 



since M is the weight of * dry steam in M -f m pounds of 
mixture. This drip pot, if perfect, would be a form of separat- 
ing calorimeter. This is not a perfect separator and so the 
quality of the steam leaving is determined by throttling the 
steam. 

The steam is throttled from the pressure shown by E to 
that shown by I. Throttling action means constant heat 
content. On expanding steam which is practically dry, it 
becomes superheated. 



I'e ^i 




q'e + x e r e = q'i + n + fc p dt = i { 


(12) 


ii - q'e 
r. 


(13) 



The thermometer H gives the number of degrees superheat 
when the saturation temperature corresponding to the pressure 
at / is found. Since the thermometer may register incorrectly 
due to the fact that the mercury column projects beyond the 
top of the thermometer well, the thermometer F is used to check 
this and determine the error. The thermometer wells are filled 
with heavy oil or with mercury and the thermometers are inter- 
changed after each reading so as to equalize errors in the ther- 
mometers. After this is done the temperature of saturation 
at the average pressure shown by E is found and the difference 
between this and the average reading of F is the error due to 
stem exposure. An error proportional to the amount of exposed 
mercury column is assumed and the average reading of H is 
corrected. From this the degrees of superheat are found, then 



THE STEAM ENGINE 181 

ii and, after this, x e is found from equation (13). The weight 
of dry steam is then 

xM 

xM 

and the actual x is X = yjt~\ (14) 

Part of this X is due to radiation from the instrument and to 
find this correction the instrument is operated with steam slightly 
superheated or dry steam and the apparent quality, x dry , is 
found. 

1 %dry 

is the correction to be applied for this instrumental condensa- 
tion. A common way to run this dry test, as it is called, is to 
attach the calorimeter to a boiler with banked fires and a shut 
stop valve in which case the steam in the steam space is dry. 

Thermometers are sometimes calibrated to read correctly when 
immersed to a certain depth and if such are used there is no need 
for the upper thermometer. Such thermometers are called 
constant immersion thermometers. 

Suppose the averages of a number of readings taken at five- 
minute intervals are as given below and the quality is desired. 

pi = 112-lb. gauge 

p 2 =2 in. Hg. 

Error in gauge = + 2 lbs. 

Barometer =29.8 in. 

T x = 335.20° F. 

T 2 = 265.2° F. 

Wt. of drip = 0.2 lbs. per 10 min. 

Area of hole = 0.01 sq. in. 

Exposed column = 120° F. for 7\. 



100° F. for T 



Observed pi =112 lbs. 

Error = 2 

Corrected p\ =110 lbs. 

Barometer = 14.64 

Absolute p x = 114.64 

Ti sat. = 337.92° F. 

7\ observed = 335.2° F. 

Error 7\ = 2.72° F. 



182 HEAT ENGINEERING 

Error T 2 = 2.72 X ^ = 2.27° F. 

Observed T 2 = 265.2° F. 

Corrected T 2 = 267.47° F. 

Observed p 2 = 2 in. Hg. = 0.98 lbs. 

Barometer = 14.64 

Absolute p 2 = 15.62 lbs. 

Temp. sat. = 215.07° F. 

Corrected T 2 = 267.47° F. 

Degrees of superheat = , 52 . 40° F. 

i for 15.62 lbs. and 52.40° F. superheat = 1177 B.t.u (from 
Marks & Davis) 

q' for 114.64 lbs. = 308.7 

r for 114.64 lbs. = 880 

308.7 + 880 x = 1177 

'-It- - 988 

m = 0.2 lbs. in 10 min. 

M = 114.64 7 X0.01 X60X10==983 

0.986X9.83 
X ~ 9.83 + 0.2 " °' 968 

The object of the drip pot is to care for any large amount of 
moisture for, if the amount of moisture is over 4 per cent., the 
equation 

q' + xt = ii 

would give an ii so small that the steam could not be super- 
heated and, of course, the moisture in the low-pressure steam 
could not be known. Hence the upper x could not be found. 
The maximum amount of moisture possible with steam super- 
heated on the lower side of the orifice depends on the pressure 
but 4 per cent, is the usual amount for pressures used in practice. 
Wherever the thermometer I shows 212° F. the drip pot is needed 
as there is too much moisture present. 

The electric calorimeter uses an electric current to dry the 
steam and by calibrating the same the amount of power for each 
per cent, of moisture at a given pressure can be found and from 
this the quality for the actual power is determined. 



THE STEAM ENGINE 183 

A condenser is not always applicable for the determination 
of the weight of steam supplied to an engine and in that case 
the engine is supplied from a boiler or group of boilers which have 
been blanked off from the others. In this way the weight of 
boiler feed gives the steam supplied to the engine if the weight of 
water left in the boiler at all times is constant. This condition 
is approximated by keeping the level of the water in the boiler 
gauge constant but even in such a case the change of temperature 
or of the rapidity of firing may make this an uncertain guide. 
To eliminate the effect of this uncertainty such tests should 
extend over a considerable time, say 4 to 6 hours and in 
some cases 24 hours. If a condenser is used a test of an hour or 
less is sufficient for accurate results after the engine is brought 
to a uniform condition. 

ANALYSES OF TESTS 

To study the action of the cylinder walls two methods of 
analyzing these test results will be examined, one analytical, 
the other graphical. 

Hirn's Analysis. — Hirn's analysis of the cylinder performance 
of steam engines consists in operating a test for a certain length 
of time and from the observations computing the actual perform- 
ance of the engine. To better illustrate this analysis a case will 
be computed. From the test the following average results are 
found : 



Time of test 60 min. 

Size of engine • 10 X 15 

Clearance 10 per cent. 

No. of revolutions during test 14,400 

Pounds of steam used 2356 

Average weight of steam per card, 9 v 14400 = 0-°816 lbs. 

Average pressure at throttle 105 lbs. 

Barometric pressure 14. 5 lbs. 

Average of quality of steam . 99 • 

Average temperature of condensate 120° F. 

Average temperature of water leaving condenser . 105° F. 

Average temperature of water entering condenser . 85° F. 

Weight of condensing water used 94,240 lbs. 

Weight of water per pound of steam 40 lbs. 



184 



HEAT ENGINEERING 



Average results from indicator cards 

Point of admission at 1 0.0 per cent. 

Point of cut-off at 2 25 per cent. 

Point of release at 3 90 per cent. 

Point of compression at 4 22 per cent. 

Abs. pressure at 1 48 lbs. per sq. in. 

Abs. pressure at 2 107 lbs. per sq. in. 

Abs. pressure at 3 37 lbs. per sq. in. 

Abs. pressure at 4 16 lbs. per sq. in. 

Work area during admission, Wa = 5-1-2-6 2.83 sq. in. 

Work area during expansion, Wb = 6-2-3-7 3.91 sq. in. 

Work area during exhaust, Wc = (3-10-8-7) — 

(9-4-10-8) -0.98 sq. in. 

Work area during compression, Wd = 5-1-4-9.. . —0.52 sq. in. 




Fig. 79. — Average indicator card from test for Hirn's analysis. 



The results above have all been averaged from the various 
readings and cards. The head end and crank end have been 
averaged together although at times they are worked up sepa- 
rately the weight of steam being divided between the two ends 
in proportion to the total volume at the point of cut-off although 
there is no true foundation for this method. Another method 
of division may be used as shown by Clayton. See page 210. 

The first point to compute in Hirn's analysis is the weight of 
clearance steam per card, M . At the point 4 steam is assumed 
to be dry. This has been indicated by experiments as far as 
they have been tried although these experiments were difficult 



THE STEAM ENGINE 185 

to perform and are not so reliable as would be desired. With 
this assumption 

7 4 = M v'\ 

or M = ^ (15) 

V 4 

V ± = total volume at 4 

v'\ = specific volume of dry steam at 4 

v'\ is found from the steam tables. After this the weight of 
the working steam per card is found. 

M = weight (16) 

2 X No. of revolutions 

Knowing M and M the quality of the steam at the various 
events of the cycle may be found if the pressures and volumes 
at these points are known from the cards. 

(17) 

(18) 

(19) 

From the pressures the values of the heat of the liquid and 
the internal heat of vaporization may be found and from these 
the intrinsic energy at all points may be computed. 

AU X = M (q\ + x m ) (20) 

AU 2 = {M + M ){q' 2 + x 2 p 2 ) (21) 

AU 3 = (M + M )(q' 3 + x 3Pi ) (22) 

AU* = M (q\ + P4 ) (23) 

From the indicator cards the external works during the 
various periods may be found. 

Work of admission = W a = 5126 (24) 

Work of expansion = W b = 6237 (25) 

Work of exhaust = W c = - 94108 + 73108 (26) 

Work of compression = W d = — 5149 (27) 

The energy added to the engine from the outside during 
admission is 

Q 1 = M(q' + xr) (28) 



Xi 


~ M v\ 




V 2 


x 2 


~ (M + M )v\ 




V, 


Xs 


" (M + M )v% 



186 HEAT ENGINEERING 

The energy removed is 

Q 2 = - M[q' + G{q' a - q\) (29) 

G = weight of condensing water per pound of steam. 

M = weight of steam per card. 

q', x, and r are for conditions of steam at the throttle valve. 

q' = heat of liquid at temperature of condensed steam. 

q' d = heat of liquid at temperature of discharge condensing 

water. 
q'i = heat of liquid at temperature of inlet condensing water. 

Now the heat added during any event is found by 

Q = A[U 2 - U 1 + W] (30) 

If the heat coming from the cylinder walls during the different 
events is represented by Q a , Qb, Qc or Q d , using the equation 
above, the various quantities are given by the equations 

Ql + Qa = AU 2 - AU 1 + AW a (31) 

Qa = - Qi + AU 2 - AU! + AWa (32) 

Q b = AUs- AU 2 + AW b (33) 

Qc = - Q 2 + AU± - AU 3 + AW C (34) 

Q d = AU 1 - AU± + AW d (35) 

If these result in positive quantities heat is given up by the 
walls while negative values mean that heat is given to the cylinder 
walls. If these are added together the net amount must be 
equal to the amount given or taken by the cylinder walls. If 
there is no source of heat in the cylinder walls they could not 
give heat so that the sum could not be positive. If negative this 
heat would finally melt the iron if abstracted for a considerable 
time. Since the walls do not change in temperature this negative 
sum must equal the heat radiated from the cylinder and 

Qa+Q b +Qc+Qd= Qr (36) 

A check equation for Q r may be determined by blocking the 
engine, filling the cylinder with boiler steam and measuring the 
condensation. Then 

Q r = M r (q' + xr- q' ) (37) 

If there is a steam jacket there could be a positive sum as 
heat could then be given up by the walls from the heat in the 
jacket. In this case 

Qa + Q b + Qc + Q d + Qi = Qr (38) 



THE STEAM ENGINE 



187 



and Qj is determined by finding the weight of steam condensed 
in the jacket while the engine is running. Mj is the condensa- 
tion due to the heat given to the steam within and 



Qi = M,- (q' + xr- q' ) 



(39) 



Q r must be determined while the engine is at rest by the con- 
densation in the jacket. 

This analysis will give the quantities Q a , Q b , Q c and Q d , but it 
will not give the value of any of them for a portion of an event. 
It will tell that so much heat has been given or received during 
an event but it will not indicate at what point this or any part 
of it has been given up. If the value of 



V = (M + M ) xl v'\ 



(40) 



be found at cut-off and for several other pressures the saturation 
curve may be drawn on Fig. 79. The ratio of the volume on 
the expansion line to that on this line for any pressure will give 
the quality x. This quality is sometimes plotted on the card as 
shown in Fig. 79. 

These equations will now be applied to the test data. The 
engine is 10 in. in diameter and of 15-in. stroke. The volume 
swept out in one stroke is therefore 0.682 cu. ft. and if the 
card is 4 in. long and has been drawn with a 40-lb. spring 
the scale of area is 



area scale = 



40 X 144 _ 0.682 



778 



X 



1.260 B.t.u. per sq. in. 



If 10 per cent, of the card length be added to each event expressed 
in per cent, of the stroke to give the total percentage volume, 
these when multiplied by the displacement will give the total 
volume in cubic feet at each event. A table is then made for 
the events as shown below, using Peabody's tables. 



Event 


Actual volume 


Pressure 


Specific 
volume 


a' 


p' 


1 


0.068 


48 lbs. 


8.84 


247.8 


846.2 


2 


0.238 


107 lbs. 


4.16 


303.5 


801.5 


3 


0.682 


37 lbs. 


11.29 


231.6 


858.6 


4 


0.220 


16 lbs. 


24.74 


184.6 


893.8 



188 HEAT ENGINEERING 

990 

M " = M§i = 00089 lbs - 

x °^?8 865 

1 ~~ 0.0089 X 8.84 ~ U ' 8D0 

9*38 

X2 = (0.0816+0.0089)4.16 = °' 633 

0^682 

* 3 " 0.0905 X 11.29 " abb * 

tfi = 0.0089 [247.8 + 0.865 X 846.2] = 8.72 

U 2 = 0.0905 [303.5 + 0.633 X 801.5] = 73.50 

U 3 = 0.0905 [231.6 + 0.668 X 858.6] = 72.80 

U A = 0.0089 [184.6 + 893.8] = 9.60 

AW a = 2.83 X 1.260 = 3.57 B.t.u. 

AW b = 3.91 X 1.260 = 4.92 B.t.u. 

AW C = - 0.98 X 1.260 = - 1.23 B.t.u. 

AW d = - 0.52 X 1.260 = - 0.65 B.t.u. 

Q l = 0.0816 [312.0 + 0.99 X 877.1] = 96.5 B.t.u. 

Q 2 = - 0.0816 [88 + 40 (73.0 - 53.1)] = - 72.0 B.t.u. 

Q a = - 96.5 + 73.5 - 8.72 + 3.57 = - 28.15 B.t.u. 

Q b = 72.80 - 73.5 + 4.92 = + 4.22 B.t.u. 

Q c = 72.0 + 9.60 - 72.8 - 1.23 = + 7.57 B.t.u. 

Q d = 8.72 - 9.60 - 0.65 = - 1.53 B.t.u. 



Q r = - 16.89 B.t.u. 

Total work 6.61 B.t.u. 

Heat supplied above 32° F 96.5 B.t.u. 

Heat supplied above temperature of exhaust 81 .3 B.t.u. 

Heat removed by walls during events: 

Admission 28. 15 B.t.u. 

Expansion —4.22 B.t.u. 

Exhaust -7.57 B.t.u. 

Compression 1 . 53 B.t.u. 

TEMPERATURE ENTROPY ANALYSIS 

The graphical temperature entropy analysis is to be used 
next and in this the same test is made but the only observations 
necessary are: 

Size of engine 10 in. X 15 in. 

Average pressure 119 . 5 lbs. per sq. in. abs. 

Average quality . 99 



THE STEAM ENGINE 189 

Weight of steam 2356 lbs. 

Revolutions 14,400 

Clearance 10 per cent. 

Average compression 22 per cent. 

Average indicator card Fig. 81 

This method has been developed by a number of persons. 
Prof. Boulvin's method has been combined with that of Reeves 
in reducing the method given below. 

The average indicator card is constructed by averaging the 
pressures at ten or more equidistant points along the card or a 



■ ii™££foe 

Temperature •> 


Pi 
O 

P 


\ 

Temperature in 


10 V 5 * 
Cu. Ft. Volume in Cu. Ft. 



Deg. F 

Fig. 80. — Temperature entropy diagram for analysis of engine performance. 

card may be selected on which the m.e.p. is equal to the average 
m.e.p. of the test. M and M are then found. 

M = -£- = 0.0089 as before. 

V 4 



M 



Wt 



2 X rev. 



= 0.0816 lbs. as before. 



The chart for the Temperature Entropy analysis is now to be 
drawn. This consists of four quadrants, Fig. 80, a p-v quadrant 
having a line for 1 lb. of saturated steam, a T-p quadrant having 
the line of saturated steam, a T-s quadrant having the liquid 
line for 1 lb. of water and the saturation line for 1 lb. of steam 



190 



HEAT ENGINEERING 



and a V-s quadrant which is used for transferring the values of 
x. This figure is constructed by aid of the steam tables and 
takes the form shown in Fig. 80. 

To transfer the indicator card to this figure, a number of 
points on the mean diagram, Fig. 81, are taken and the heights 
from the line of absolute zero pressure and lengths from absolute 
zero of volume are measured and tabulated. The lengths are 
then multiplied by the scale of volume of Fig. 81 and divided by 
the product of the scale of Fig. 80 and the value of M + M 
in order to get the distance to each point if 1 lb. of steam were 
present on the expansion line. The heights are multiplied by 
the scale of 81 and divided by the scale of 80. Thus, in the 




V 16 






c 




\15 e 








\ 10 


\^14 


13 


12 




d\ 










! 



Fig. 81. — Average indicator card from test arranged for T-S analysis. 

problem considered there is a 4-in. card for a 10 X 15 engine 
with a 40-lb. spring and 

M + M = 0.0816 + 0.0089 = 0.0905 

Fig. 80 has scales of 2.5 cu. ft. to the inch and 20 lbs. per square 
inch to the inch. The cylinder has a displacement of 0.682 cu. ft. 
with a 4-in. card. The lengths are therefore multiplied by 

^X^-X--0754 
4 X 0.0905 X 2.5 U *^ 4 



in order to get the length to lay off in Fig. 80 so as to represent 
the card for 1 lb. of steam on the expansion line. The multi- 
plier for height is merely 40/20 = 2.00. These distances are 
tabulated to aid in the work as shown. 



THE STEAM ENGINE 



191 



00 


Original 
card 


Analysis 
card 


to 

d 
"o 


Original 
card 


Analysis 
card 


a 
'3 

P4 


33 


<S ID 
O . o 

a a fi a 
Q- 5 .2 


1.2 
§.2 


8 g 

.2 a& °3 
Q -rt -2 


1.2 

3-s 


2 fl 2 § 

P ,2 
u 


+3 . 

-d a 
W — ■ 


S g 

o 


1 


2.85 


0.40 


5.70 


0.30 


11 


0.40 


4.4 


0.80 


3.31 


2 


2.82 


0.90 


5.62 


0.68 


12 


0.40 


3.4 


0.80 


2.56 


3 


2.78 


1.23 


5.56 


0.93 


13 


0.40 


2.4 


0.80 


1.81 


4 


2.64 


1.40 


5.28 


1.05 


14 


0.40 


1.2 


0.80 


0.93 


5 


2.01 


1.90 


4.02 


1.43 


15 


0.51 


0.9 


1.02 


0.68 


6 


1.56 


2.40 


3.12 


1.81 


16 


0.80 


0.6 


1.60 


0.45 


7 


1.28 


2.90 


2.56 


2.18 


17 


1.18 


0.4 


2.36 


0.31 


8 


1.10 


3.40 


2.20 


2.56 


18 


1.50 


0.4 


3.00 


0.31 


9 


0.93 


4.00 


1.86 


3.00 


19 


2.00 


0.4 


4.00 


0.31 


10 


0.70 


4.20 


1.40 


3.16 


20 


2.50 


0.4 


5.00 


0.31 





^ 




¥e 


— Z \ 


o 




f 




r- ^~_— ' 




>> / 




1 — 




p. / 










O / 






-L . 




S / 






\£ 


5 21 


7 






a 


~-"^^/ 








Temperature 








\ 


® 


inl 








£3 


""A*! 




































\ 


Ot 


20 v\ 

19 Vs 






\ 




18 \\l ,/ 


</X 


5 ^yv 




\ V^V^ 




A 16 A ^^ 


\. 


5\ \ 10 \*^ 


\ 


Si fix, . . , Vll — 


£ 14 13 12 ^"* 


Temperature 


Volume 



Fig. 82. — Construction of T-S diagram from p-v diagram. 

This method considers only 1 lb. of steam to be present on 
the expansion line. 

The p-v diagram is now laid out from the table giving the fig- 
ure shown. This is then transferred from the p-v to the T-s 
quadrant by drawing a constant pressure line until it strikes the 



192 HEAT ENGINEERING 

saturation line in the T-p quadrant and this fixes the tempera- 
ture of the point A. A vertical line from this temperature fixes 
a line in the T-s quadrant on which the point will lie. It would 
be at a if it had a quality of zero and at b if of quality 1, and in 
the p-v quadrant it would be at c if of zero quality and at d if 
of unit quality. 

Now v — v f = xv" (41) 

and s - s' = y ( 42 ) 

In other words the volume change and the entropy change from 
liquid to steam depends on x. Hence the point A' in the T-s 
quadrant must divide a-b in the same way that A in the p-v 
quadrant bisects c-d. To construct such a point the point of 
intersection e of a horizontal from b and a vertical from d is 
joined to the intersection/ of a horizontal from a and a vertical 
from c. The line ef is then the quality transfer line for if a 
vertical is drawn from A to this line intersecting it in h and then 
a horizontal is drawn from h to ab, this will fix the point A' . 
The same pressure line and construction transfers B to B' . If 
this is carried on in the same manner for all points a figure such 
as shown in the T-s quadrant, or enlarged in Fig. 83, is found in 
the T-s quadrant corresponding to the p-v diagram. The scales 
to which the p-v quadrant has been drawn were 

1 in. = 2.5 cu. ft. 

1 in. = 20 lbs. per sq. in. 

The T-s quadrant was drawn with 

1 in. = 0.25 units of entropy 
1 in. = 25° F. 

The area scales are: 

1 . 2.5 X 20 X 144 no , -p + , , 
1 sq. m. = =^z = 9.25 B.t.u. {p-v) 

1 sq. in. = 25 X 0.25 = 6.25 B.t.u. (T-s) 

The area of the indicator card on the p-v plane was 7.94 sq. 
in. and on the T-s plane the area was 11.76 sq. in. 

These each reduce to 73.5 B.t.u. for the area. 

The T-s plane for Fig. 83 has been turned through 90°. 
This is the conventional T-s diagram for the indicator card. 



THE STEAM ENGINE 



193 



It is now necessary to add other lines to the T-s diagram in 
order to interpret it. Letters have been added to the diagram 
at critical points of Fig. 83. The diagram is for 1 lb. of total 
steam and this is the amount present on be only hence the figure is 
true only for this short line. From the point of admission / to 
b there is a changing weight of steam and from c to e there is 
a changing weight. The points on these lines are only drawn 
according to the actual volumes which are proportional to en- 
tropies because each is proportional to Mx. To study the line 

Mo 
ef on which M ,° M pounds are present, the line corresponding 

to zero entropy change must be drawn. If AC be drawn so 
that 

M 0.0816 = QM 



E B M + M c 



0.0905 




Zero Line fa I 

at Proper 
Distance 



Fig. 83. — T-S diagram of indicator card. 



then line AC would be the liquid line for the boiler steam only, 
while the distance CB or the distance between AC and A r B repre- 
sents entropy of the liquid for the clearance steam. 

The distance eF represents the entropy of the clearance steam 
at the beginning of compression and GH is laid off at this distance 
from CFA. This line would be the line of compression if the 



13 



194 HEAT ENGINEERING 

clearance steam did not change in entropy because the entropy 
of the liquid CB for this steam plus the entropy of vaporization 
BH is constant. GH, although not an adiabatic on this figure, 
is the line of compression if the clearance steam were com- 
pressed without giving up heat. The line of compression is 
seen to pass to the left of the line GH which means that heat is 
taken up by the cylinder walls. The amount of heat is found 
by drawing fg parallel to GH to the back pressure line ed and 
then dropping the perpendiculars to h and i at absolute zero of 
temperature getting the area hgfei as the heat removed in com- 
pression. To find the heat lost during admission from f to b 
it is necessary to find the position which should have been occu- 
pied by the steam entering from the boiler if no heat had been 
taken from the clearance steam or the steam entering the en- 
gine. It is necessary to mark the line corresponding to the pres- 
sure at the throttle valve and draw this on the diagram. This 
will be the line ECBH. On this line EC is the entropy of the 
liquid of the boiler steam, CH is the entropy of the clearance 
steam and in addition to this there must be HJ for the vaporiza- 
tion of the steam from the boiler. If x is the quality of this 
steam the distance HJ is equal to 

*wfwS BI) (44) 

T 

Since BI is the value of ™ for 1 lb. of total steam, 
BI = BH + HI 

The point which should have been at J is now at b and the area 
nJlk is called the loss due to initial condensation. Since the 
reference line to care for other losses is shifted to gM the point 
J is moved to K and the loss due to initial condensation is 
nKtk. The area mnba is called the loss due to wiredraw- 
ing. There is still a further loss jMmaj which occurs during the 
early part of admission. 

The area prbk is the loss to the cylinder walls after cut-off 
and area prc'o' represents a gain from the cylinder walls, and 
occ'o' represents a loss during the last part of expansion. 

ksc'co — brs = net gain from cylinder wails during expansion. 

iedco = heat removed during exhaust and dcu = loss from 
incomplete expansion. 



THE STEAM ENGINE 



195 



Now the line GH is a liquid line for the cylinder feed and 
hence ieHJl = the heat supplied with the cylinder feed for one 
pound of total steam, above the temperature of the exhaust. 

The work is efabcd 

Hence 
ieHJl + (ksc'co — srb) — abcdef — iedco — hgfei = heat removed 
during admission. 

This will be found to be equal to the area of fMKtkbaf if gf 
is continued to M and JK is made equal to MR. 

The various losses are given as follows (numerical values 
should check with Hirn's analysis) : 



Area 



Sq- in. 



Per cent, 
of work 
on T-s 



B.t.u. 



Heat supplied above exhaust = ieHJl or hgMKt. 

Work = efabcd 

Loss during first admission = fMma 

Loss during throttling = amnb 

Loss due to condensation = nKtk 

Heat added from walls during expansion = 

ksc'co — brs 

Loss due to free expansion = dcu 

Heat necessarily removed during exhaust = ieuo . 
Heat removed during compression = hgfi 



144.98 

11.76 

0.71 

0.30 

44.32 



8.43 
4.74 



1232.00 

100.00 

6.04 

2.55 

376.20 

23.80 

13.95 

717.50 

40.40 



905 . 00 

73.50 

4.45 

1.88 
277.00 

17.50 

10.30 

527.00 

29.60 



The quality of the steam at cut-off is given on Fig. 83 by the 
ratio 

(45) 



X = 



wv 44.6 



= 0.645 



and this can be found in the same manner for any point of the 
expansion curve. For the compression curve x may be found by 
drawing from e a line eO so that its distance from the line AC 

M c 



is 



M + M c 



times the distance from A'B to IN. Thus 



BO = BI 



M c 



M + Mc 



If the ratio of the distance from AC to the compression line 
to the distance from AC to eO at the same level be found this 
ratio is x. AC, A'B and eO make an entropy diagram for the 
compression steam with a bent axis AC. 

As before mentioned the only true lines from which x can 



196 HEAT ENGINEERING 

be found are be and ef. The other lines are obtained by following 
a scheme in which the product Mx at any point could be found 
but not either of the terms of the product. Hence one cannot 
tell what losses occur at various points on this line. There should 
be an agreement by the two analyses. 

MISSING QUANTITY AND INITIAL CONDENSATION 

Steam of quality 0.99 was admitted into a cylinder and after 
entering the cylinder it was found that the quality at cut-off has 
been reduced to 0.633. In other words 37 per cent, of the mix- 
ture is moisture. This has been caused by the action of the 
cylinder walls. Fig. 84 shows the actual form taken by the 
indicator card of an engine. The events in most cases take 

place slowly giving rounded 
corners; there is a drop due 
to throttling on the steam 
line and release occurs before 
the end of the stroke. 

If the weight of dry steam 
shown by the cards, Fig. 84, 
is found by the formulae be- 
low and this is subtracted 

from the amount of steam 
Jig. 84. — Actual indicator card from 

engine. actually supplied, the differ- 

ence is called the "missing 
quantity." It is expressed as a percentage of the steam actually 
supplied. It is caused by initial condensation of the steam. 

ft r i i i 

^ = zr[ al X ^ - %^J < 46) 

M a = apparent weight from diagram. 
F = area of piston in square feet. 
L = length of stroke in feet. 
D = length of card in inches. 

I = per cent, clearance from Fig. 84. 
v" = specific volume of dry steam. 
ait a 2 = distances from clearance lines to points in inches. 

If M equals the real weight taken per card, 

M - M a M m 
—ji^ = -^- = m.q., missing quantity. 









ID 


1 ^\ 




V* 




— a 2 — *\ 





THE STEAM ENGINE 197 

This is almost equal to 1 — x 2 from either of the previous 
analyses. 

This initial condensation is due to the effect of the cylinder 
walls. As the steam in the cylinder drops in pressure its tem- 
perature changes and the moisture on the walls of the cylinder 
from condensation is evaporated removing heat from the metal 
and moisture and cooling them. During the exhaust this action 
continues, the walls being so cooled that when fresh steam enters 
the cylinder some of the steam is condensed and settles on the 
walls of the cylinder. This enables the walls to absorb heat 
from the steam at a faster rate than it would were dry vapor in 
contact with the metal. This condensation produces the miss- 
ing quantity and the presence of a film on the walls of the cylinder 
seems to make it easier to carry heat to the walls during admis- 
sion while during the exhaust the possibility of evaporating this 
water means that during this time much heat is given up by the 
walls and discharged with the exhaust steam when it can be of 
no value for driving the engine. 

If superheated steam be furnished, it will give up its heat to 
the cylinder walls but the absence of the moisture film makes 
this action slower and if the superheat is sufficient the necessary 
heat may be abstracted before the steam is reduced to the 
saturated condition or before much condensation occurs. At 
release the absence of considerable moisture in the cylinder 
prevents the removal of much heat during the exhaust. Thus 
the superheated steam prevents the excessive abstraction of 
useful heat by the metal during admission and its restoration 
when it cannot be used. It is for this reason that superheated 
steam is such a valuable medium to use. This accomplishes a 
greater increase in efficiency than that which is shown by theo- 
retical considerations. 

To determine the amount of steam used by an engine it 
will be necessary to know this missing quantity. Then 

M a 



M = 



1 — m.q. 



To find the value of m.q. theoretical and empirical formulae 
have been proposed. 

It will be evident that the amount of condensation per stroke 
must depend on the surface exposed to the steam at cut-off, 
the temperature range and the time during which this surface 



198 HEAT ENGINEERING 

is exposed to the steam or inversely on the number of times 
it is exposed to the action of the steam per minute. There 
have been several formulae proposed. Some are quite complex 
taking into account many variables. The more complex, al- 
though they may be correct in theory, are so changed by slight 
changes in conditions that the possibility of error is as great as 
in the less complicated formulae. Some of the formulae pro- 
posed are as follows: 

m.q. 217(r - 0.7) _ 

T^h ' ~^VW (Perry) (47) 

m.q. 15(1 + r) ,_ , 

r4 = ~ivw- (Perry) (47a) 

m.q. 100 loqr t „ t % 

r^fe = nvt (Cotterlll) (48) 

m. qi _ = 30VT (Thurgton) (4Q) 

1 - m.q. dy/N 

100 It 
m.q. = iQfk I t (Callendar and Nicolson) (50) 

100 m.?. = 1Mr d x? ^ ( Rice ) ( 51 > 



Pressure 


a 


P 


a 


60 


0.568 


0.412 


133 


80 


0.517 


0.384 


106 


100 


0.466 


0.359 


87 


120 


0.414 


0.333 


78 


140 


0.363 


0.306 


75 



In the above formulae, 

r = ratio of expansion, the volume at end of 
stroke divided by the volume at cut-off. 

d = diameter in inches. 
pi = absolute steam pressure in lbs. per sq. in. 
N = r.p.m. 

I = percentage clearance. 

m.q. = c\[T h - T e ] d cos" 1 ^^-^ + [T b - T e ] 

(127e + 0.055)s } ^ (Marks) (52) 

c = 0.02. 
T b = abs. temp, at cut-off. 



THE STEAM ENGINE 



199 



m.q. 



A 



T e = abs. temp, during exhaust. 
d = diam. in feet. 
e = fraction of volume at cut-off. 
s = stroke in feet. 
b = fraction of volume at compression. 

sT 



(Heck) 



(53) 



0.27 
VW\ep 

77i. q. = missing quantity for small compression or 
(1 — x) at cut-off for large compression. 
N = r.p.m. 
s = nominal cylinder surface divided by volume, 

each in feet, 



■.-G4- 



T = temperature range from limiting pressures 

on curve Fig. 85. (T 1 -T 2 .) 
p = abs. pressure at cut-off in lbs. per sq. in. 
e = ratio of total volume at cut-off to volume 

swept out by piston. 



u 175 

o 

% 150 

M 

O 

1 125 
















/ 


r 
















/ 




3 
I- 

£ alOO 














/ 


/ 




O 3 

a a 

o 

a 50 














/ 
























a 

5 25 

9 

1 







































50 100 150 200 250 300 350 400 450 
Values of T 

Fig. 85. — Heck's value of T for different pressure. 

The basis for the form of these formulae may be given as 
follows : 

The surface exposed to the steam at cut-off in square feet is 



2— * + Trd- r L+F p = F C 



(54) 



d = diam. in feet. 



200 HEAT ENGINEERING 

L = stroke in feet. 
- = point of cut-off. 
F p = surface of passages. 
The time during which this is exposed is 

k -^ minutes 

where k = a fraction less than 1 

N = r.p.m. 

The temperature range for this surface is some function of 
T. It is not equal to the actual difference in the temperatures 
at cut-off and compression as the range in temperature in the 
metal is much less than this. The quantity of steam necessary 
to give the heat per stroke required by the cylinder walls will 
be proportional to the surface, time and temperature range, or 

M a = ¥F e ^f(T) (55) 

To express this as a fraction, m.q., of the steam shown by the 
indicator card it must be divided by the steam shown by the card. 
This steam is given by 



* 



M a = 77 compression steam. (56) 

k" 
Now the specific volume v'\ — — approximately since the 

curve of saturated steam on the p-v plane is nearly a rectangular 
hyperbola. (See Fig. 96.) 

M a = fc'"£ ^L(approx.) 

Hence 

M s m.q. . 1V 1 . f m\Fc r 

mr = ^ — = m.q. (approx). = h i^f{T)—^ (57) 

M a 1 — m.q. * v PF J N J v J ird 2 p v J 

Now 7 2 c may be called s. F c is practically equal to 2 -j — (- 



irdL, so that s = j + -, 



THE STEAM ENGINE 201 



This is a function of -? giving 

m.q. = *- J2 /(r) ori'Il /(D (58) 

If r is written as -, e, being the relative total volume at cut-off 
c 

this reduces to 

m - q - = F J /(r) (59) 

By examining a number of tests of all sorts of engines and with 
different speeds, pressures and cut-offs for the same engine, 
investigators have been led to the empirical forms shown above. 
What in the simple theory appears as the first power has in 
general been changed to a fractional power. In all of the simpler 
formulae the form agrees in the positions of the terms with the 

k 
simple theoretical form shown. In place of -^ being used for the 

time term — 1= or , , — has been used. Heck assumes that the 

Vn Vn 

last two terms of F c are equal to wdl, thus making F c equal to the 
inside surface of the piston -displacement, 

<d 



f. = *-d(| + i) 



From evidence shown by those who have derived these for- 
mulae it appears that the formula of Heck 

0.27 

Vn Mpe 



h VS (57) 



is a good one to use. It is applicable to non-jacketed engines 
when supplied with dry steam. For this reason it cannot be 
used when the supply is superheated steam nor for the lower 
cylinders of multiple expansion engines when the steam supply 
is quite wet. If, however, this steam is dried by a separator or 
by reheater coils before entering the other cylinder, it may be 
used. For jacketed engines it might be used by deducting the 
surfaces heated by the jacket from the area F c in finding this or 
the quantity s. 

This formula shows that the proportional amount of condensed 
steam varies inversely as the square root of the cut-off getting 
less as the cut-off becomes greater. This must not be confused 



202 



HEAT ENGINEERING 



with the actual amount of steam condensed which will not vary 
very much as the cut-off changes. The large part of the conden- 
sation takes place when the steam first enters the cylinder and 
as the piston moves along the additional amount is not great. 
The surfaces which are most important in the condensation of 
steam are the cylinder head and piston. For this reason greater 
gain is to be expected from the jacketing of the heads than from 
the jacketing of the barrel of the cylinder. It might pay to 
supply the hollow piston with steam, making a jacket of the cored 
spaces. The value of s decreases as the size of the cylinder in- 
creases giving less condensation in large cylinders than that 
found in small ones. 

This is to be applied to the card below, Fig. 86, which is the 
assumed card in the design of a 20-in. X 24-in. engine to run at 80 
r.p.m. The clearance is 7 per cent.; cut-off is at 25 per cent. 
Pressure at cut-off is 125 lbs. abs. and back pressure 17 lbs. abs. 



24^20 



= 1+2.4= 3.4 



12 12 

^I^JFrom curve, Fig. 85. 

T = 360 - 217 = 143 
e = 0.25 + 0.07 = 0.32 
°-27 ISA X 143 

m - q - - m VMtS = °- 219 




Fig. 86. — Card for proposed 20 X 40 engine. 
Amount of steam indicated at cut-off is given by 

X 10 2 
0.32 X 4.36 



Mi = (0.25 + 0.07) p^ > 



3.581 



12 144 

= 0.390 lbs. 



V'l25 



THE STEAM ENGINE 203 

a + * • 0.33 X 4.36 ft _ A1 _ 
Steam at compression = ~o oo — = 0.0515 

Indicated steam per hour = (0.390 - 0.0615) X 80 X 2 X 60 
= 3160 lbs. 

Probable steam per hour ^ _ o 1Q = 4000 lbs. 

Mean height = 1.41 in. 
m.e.p. = 56.4 

56.4 X H X 314 X 80 

H 'P- = 3300(r - X 2 = 172 

Probable steam consumption, 

Steam cons. = 179 = 23.2 

This result is low so that the missing quantity will be com- 
puted by some of the other formulae. 
By Perry's formula (48) 

15 r i.o7] 

ra.g. = L °- 32 J = o 37 
1 - m.q. 20 X V80 
m.q. = 0.271 

By Thurston's formula 

30* 

1 - m.q. 20 X V80 
m.q. = 0.237 

By Cotterill's formula 

1.07 
100 log t™ 

m -V- _ ^ = 0.293 

1 - m.q. 20 X V80 

m.q. = 0.226 

The result of 23.2 lbs. per hour per horse-power was low for 
this type of engine so that Heck's formula would not give as good 
results as Thurston's, Perry's and Cotterill's. If 0.30 is taken 
for the value of mq. The steam consumption is equal to 

24.3 X * ~ p g^ 9 - 27.2 lbs. 



204 HEAT ENGINEERING 

On account of valve leakage this result would be increased 
slightly. 

EXPERIMENTS OF EFFECT OF CYLINDER WALLS 

This influence of the cylinder walls has been experimentally 
studied by a number of investigators. Callendar and Nicolson, 
in their paper presented in the "Proceedings of the Institution 
of Civil Engineers of Great Britain," Vol. cxxxi, gave results 
obtained by them on a 10% X 12 Robb engine in 1895 with a 
flat slide valve behind a pressure plate. The piston displace- 
ment was 0.601 cu. ft. and the clearance volume was 0.060 cu. 
ft. The engine was made single-acting so as to study the action 
in a better manner. To find the temperature of the metal in 
the cylinder head eight holes were drilled at regular intervals at 
1)4 in. from the center of the head but extending to different 
depths. The thicknesses of metal remaining at the bottom of 
seven of these holes were 0.01 in., 0.02 in., 0.04 in., 0.08 in., 0.16 
in., 0.32 in. and 0.64 in. Along the length of the cylinder at the 
end of stroke and at 4, 6 and 12 in. from it a pair of holes were 
drilled in the side, one to within 0.04 in. of inner surface and 
the other % in. At 2, 8, 10, 14 and 16 in. from the end single 
holes were drilled leaving J£ in. at bottom. There were also 
four holes at the middle of the stroke distributed around the 
circumference of the cylinder extending to within % in. of the 
inner surface and finally three vertical holes two inches deep were 
drilled along the side of the barrel at 1 in., 7% m - an d 15 in. for 
the use of mercurial thermometers or platinum resistance ther- 
mometers, while in the other holes thermocouples were used. 

The thermocouples were made by soldering wrought-iron 
wires at the bottom of the holes. To make the cold junction, 
the same kind of wire was attached to cast-iron blocks cast from 
the same ladle as that from which the head was cast. These 
were immersed in a paraffine bath at 212° F. The potential 
was measured by a very sensitive galvanometer by balancing 
the potential on a potential wire. The formula for the voltage 
in microvolts produced by this couple was found to be given by 

E = 1692 - 17.86* + 0.0094* 2 (58) 

if the cold junction was at 100° C. and the hot junction at t°C. 
Later one of the couples in the cylinder was used as the cold 



THE STEAM ENGINE 205 

junction and the drop of temperature between these two was 
measured. 

This method was somewhat similar to that used by Prof. 
E. Hall of Harvard (Trans. A. I. E. E., 1891) but his results 
were not very extensive. 

To find the temperature of steam the authors used a platinum 
resistance thermometer in the cylinder 3 in. from the face of the 
piston and also one in a small %-in. hole in the center of the 
piston head. 

To get the temperature at a definite point in the stroke by 
any couple or platinum thermometer a pair of revolving brushes 
were attached to the shaft. One brush made contact with a 
central copper tube and the other with a sector mounted on a 
circular disc. The sector was one-thirtieth of a circumference 
in length. The disc could be rotated and by a scale and vernier 
the position of the crank for any observation could be read. A 
number of sectors on the disc reduced the amount of motion neces- 




42 45 48 51 



Fig. 87. — Indicator card marked with points at various sixtieths of a revolu- 
tion. From 20 X 40 engine. 

sary. To facilitate the change of circuit to different couples, 
mercury cups were used. 

To show the results of these tests, Fig. 88 has been constructed 
from the card, Fig. 87 by finding the saturation temperature for 
pressures of the steam at points corresponding to definite crank 
angles. This temperature from the steam tables is plotted to 
sixtieths of a revolution, giving the curve. The marks X give 
points similar to the results shown by the platinum thermometer 
while the points marked O show the temperatures of the ther- 
mometer in the steam space in the cylinder head. The curves 
illustrating the variation of temperature of the metal at 3^5 in. 
from the inside surface in the head and that at holes in the side 
at 4 in. from end are shown to a larger scale above the card. 
These curves are ideal and are drawn to indicate the results of 
Callendar and Nicolson. 



206 



HEAT ENGINEERING 



The surprising results of the actual tests by Callendar and 
Nicolson were that at Jfoo in. from the inside surface the variation 
of temperature was only 4.3° F. at 100 r.p.m., while at %5 m - 
from inside the variation was 6° F. at 46 r.p.m. and 4° F. at 73.4 
r.p.m. At }4 in. from the inside the change in temperature due 
to cyclic variation was practically zero. 



400 



S 350 



250 F. 



200 F. 



/ 






Is Si ^>^ ^Aro/, 



^ A ^V?55Kr^fcL eo ' 0on P»e«n Head _.<? 



^2§&* 



Clearance 



Cycles shown by Couples 




Area cbd + abc = 1. 

Sq. In. 
X = Temp.byPt.Ihermometer Attached to Piston 



d' 



3 



12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 
Sixtieth of One Revolution 
l"*=10-ir Rev. 



Fig. 88. — Diagram of temperature at various positions of piston for dif- 
ferent movement of crank. Temperatures taken in head, cylinder wall and 
steam space. (After Callendar and Nicolson.) 



Along the length of the barrel the first hole situated in the 
clearance space and exposed to the same steam action as the 
heads showed 13.5° F. variation at }^5 in- from inside at 44 r.p.m. 
when the heads at this thickness showed 4.9° F. variation. 
This amount of 13.5° F. at J^5 in. from the surface probably corre- 
sponds to about 20° F. variation at the surface and this particular 
result was the largest obtained by the experimenters. The 



THE STEAM ENGINE 207 

variation at 4 in. was found to be 5° F. and at 6 in., 3J£° F. It was 
found that these temperatures begin to rise before the piston 
reaches these points showing that there was conduction along the 
barrel and also the steam could leak around the piston as far as 
the rings. A curious result near the middle of the stroke is the 
temperature that is shown on the inner surface which is lower than 
that some distance farther out. This shows that there is a flow 
of heat from the metal to the steam at this point. The experi- 
ments seem to show a gradient of 0.55° F. per inch in the head 
while that on the side wall of the barrel was a variable quantity 
in an axial direction being greatest at the center of the stroke 
where it was 9.3° F. per inch. 

Experiments were made to determine the conductivity of 
cast iron giving 5.5 B.t.u. per square feet per degree per hour 
for 1 in. thickness and then the diffusivity, which is the ratio of 
this conductivity to the thermal capacity of the same amount of 
metal, was computed. 

The results of these experiments showed that the effect of 
moisture in the steam is to increase the condensation while 
superheat decreases it. The percentage amount of condensa- 
tion varies inversely as the ratio of expansion making the actual 
amount of condensation practically constant. The effect of 
initial pressure is complex so that an increase does not mean 
an increase in condensation and the same may be said of the 
change in back pressure. Early compression means a shorter 
time for cooling and hence decreases the condensation. Jacket- 
ing reduces the surface which causes condensation and for that 
reason reduces the amount of condensation. 

To find the maximum condensation possible, the limiting 
amount as they called it, the average height of the temperature 
degree diagram, Fig. 88, is found and when the area above this 
line is expressed in degrees and sixtieths of a cycle it will give 
the thermal units per hour per square foot when multiplied by 
45. The surface considered for 20 per cent, cut-off on this en- 
gine which was a Corliss engine was 8.9 sq. ft. on each end. 
45 X Area (cbd + abc) X scale X surface of clearance = 45 X 
1.89 X 10 X 50 X 8.9 X 2 = Total B.t.u. of condensation per 
hour = 758,000 (59) 

This quantity is then multiplied by 

i + Vn Vn 

3 -Tvf or JTvf m 



208 HEAT ENGINEERING 

to make it correct for the speed of N revolutions per minute for 
cut-off or release respectively; while multiplying by 1.4 gives 
the result for a double-acting cylinder. 

Thus, from the Fig. 88, the following results for a 20 in. X 
42 in. engine at 80 r.p.m. : 

758,000 X I ^~\ S0 . X 1.4 = 884,000 B.t.u. per hr. 
3 + V80 

If the pressure at entrance from the card were 100 lbs. absolute 
for which r = 887.6, the steam condensed per hour will be 

_ _ 884000 Mtr „ 
= ~8876" = 

This engine uses 4000 lbs. of steam per hour so that the 
missing quantity is 0.25. 

The actual condensation may be much less than this and, if 
the mean temperature of the cylinder wall is known, the area 
above this line to any event gives the actual condensation when 
handled in the manner shown. The net area to a point beyond 
the crossing of the mean line gives the difference between the 
condensation and the reevaporation. The multiplier is different 
for cut-off and release. 

VALVE LEAKAGE 

Callendar and Nicolson found that in their engine much of 
the apparent missing quantity was due to leakage of steam 
beneath the valve into the exhaust and this steam never entered 
the cylinder. The valve leakage of steam was found to depend 
on the periphery of the valve, the lap and the pressure difference 
or 

W = Kl{Vi - ^ (61) 

s 

when pi and p 2 were the mean pressures on the two sides in 
pounds per square inch, s was the lap in inches and I was the 
periphery in inches. 

The value of this leakage term was undoubtedly large in 
the experiments under review but in many engines it may be a 
small quantity. In the engine used K was equal to 0.02. 

The platinum thermometer projecting into the steam of the 
cylinder 3 in. from the piston face indicated (Fig. 88) that 



THE STEAM ENGINE 



209 



160 Ci 




140 


Wall Temp..// fi 


120 


-- ^/ //I 


|ioo 




f 80 


Steam Temp./ / 


p. 


\/ / 


§ 60 


k— -X' 


40 


Sat. Temp. 


Corresponding 


20 


to Pressure 


0°C. 





Piston Travel 

Fig. 89.— Results of 
Duchesne on tempera- 
ture of wall and steam. 



the steam was practically saturated except during compres- 
sion when it was superheated, while that in the %-in. hole in the 
piston head was highly superheated during all of the stroke 
except at cut-off. This was undoubtedly due to the superheating 
effect of the metal when the pressure was low causing the steam 
to be superheated still further during compression. 

George Duchesne in the Revue de Mechanique 1 gives results 
of an investigation with a hyperthermom- 
eter which consisted of a multiple silver- 
platinum couple 0.002 mm. thick. This 
couple responded rapidly to the changes of 
temperature. He finds the cycle of Fig. 89 
for the wall, steam and the saturation tem- 
perature of the steam in his engine on a 
temperature-piston travel diagram. In 
this it is noted that the wall temperature 
is usually higher than the steam and that 
the steam is saturated except during com- 
pression when it is superheated, becoming 
saturated at the point where it reaches the 
wall temperature. Of course this metal 

cycle is true for one point in the cylinder but for another point 
the cycle would probably be lower. Duchesne states that the 
compression should not be carried higher than the temperature 
of the metal of the head for, if this is done, a loop is found on the 
card, Fig. 90. That this loop is due to excessive condensation 
was proven by the fact that, when air was used with this amount 
of compression, the loop was not shown, 
compression being carried up to a higher 
pressure. The loop may not be found in 
high-speed engines when the piston speed 
is faster than the speed of condensation. 

This compression, however, does not 
affect the steam consumption very greatly 
as has been shown by Prof. John Barr 2 
in 1895 and by E. Heinrich 3 in 1913. In these experiments there 
was a slight change in the steam per indicated horse-power hour 

1 Revue de Mechanique, 1897, pp. 925 and 1236. Power, June 28, 1910; 
May, 1911. 

2 Trans. A. S. M. E., 1895, p. 430. 

3 Zeit. des Verein Deutscher Ing., Vol. 58, Jan. 3, 1914. 

14 




Fig. 90. — Duchesne 
compression loop due to 
excessive compression. 



210 



HEAT ENGINEERING 




Pressure orTemperature 
of Steam vs Stroke 




Temperature of Metal 
vs Stroke 

Fig. 91. — Adams' 
results. 



with change in compression but this was very slight. Heinrich 
found that by adding plates and increasing the clearance surface, 
although the volume remained the same, the water rate was 
increased, showing the harmful effect of sur- 
face. 

In 1895, at Cornell University, Mr. E. T. 
Adams used a couple in the cylinder wall near 
the inside surface (Koo m and attached it to 
a galvanometer of short period. He photo- 
graphed the beam of light from the mirror of the 
galvanometer on a sensitive paper moved by 
the piston. This result is shown in Fig. 91, 
which shows a rapid drop at the opening of re- 
lease probably due to the evaporation of water from the surface 
and, as the heat flowed in from the outer metal of the wall, 
this temperature rose again. 

STEAM CONSUMPTION BY CLAYTON'S METHOD 

J. Paul Clayton 1 has studied the results of actual tests of engines 
and has shown that the expansion lines of steam engines and, in 
fact, the expansion lines from any engine using an elastic medium 
are polytropic curves of the form pv n = K unless there is some 
peculiarity such as leakage or some other disturbance which is 
not uniform. He then investigated 
the values of n from different en- 
gines, finding the value of n by plot- 
ting a logarithmic diagram of pres- 
sure and volume as shown for the 
cards of Fig. 92 taken from a 20 
in. X 42 in. engine at 78 r.p.m. with 
4 per cent, clearance. The table 
below shows the method of con- 
structing the diagram Fig. 93. If the clearance is properly 
measured and the correct scale of the spring is used, the ex- 
pansion lines become straight lines on the logarithmic diagram 
and the slopes of these lines are the values of n since 

log pi - log Vi 



60* Spring 




Fig. 92. — Card from engine for 
Clayton analysis. 



log Vt, — log Vi 

1 Trans. A. S. M. E., 34, p. 17. 
Bulletin 58, Engineering Experiment Station, Univ. of Illinois. 
Bulletin 65. 



(62) 



THE STEAM ENGINE 



211 



The variation of this line from a straight line, as shown in the 
figure, would indicate leakage. 



Point 


Volume in fifths 

inches from 

clearance line 


Logarithm of 

volume in fifths 

inches + 1 


Pressures in fifths 

inches from 

absolute zero 


Logarithm of 

pressure in fifths 

inches + 1 


1 


0.70 


0.846 


8.65 


1.936 


2 


2.20 


1.340 


8.40 


1.924 


3 


3.88 


1.590 


8.30 


1.919 


4 


5.50 


1.740 


8.28 


1.918 


5 


6.25 


1.795 


7.97 


1.902 


6 


7.47 


1.873 


6.50 


1.813 


7 


9.35 


1.970 


5.20 


1.716 


8 


12.80 


2.106 


3.80 


1.579 


9 


15.85 


2.200 


3.15 


1.498 


10 


18.92 


2.276 


2.05 


1.424 


11 


17.80 


2.250 


2.18 


1.338 


12 


15.20 


2.182 


1.88 


1.274 


13 


12.55 


2.099 


1.85 


1.267 


14 


9.50 


1.978 


1.80 


1.255 


15 


5.70 


1.756 


1.80 


1.255 


16 


3.52 


1.546 


' 1.75 


1.243 


17 


2.80 


1.447 


.2.00 


1.301 


18 


1.75 


1.240 


3.15 


1.498 


19 


1.25 


1.096 


4.35 


1.638 


20 


0.70 


0.846 


7.20 


1.864 



2.00 



2 1.75 



1.50 



1.00 




0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 

Logarithm of Volume 

Fig. 93. — Logarithmic diagram of steam engine indicator card. 



In preparing the table, distances to the true zero of pressure 
and of volume have been found and then the distances from these 
lines to any point have been measured in fifths of an inch. The 
logarithms of the numbers have been increased by unity to have 



212 



HEAT ENGINEERING 



all of the numbers positive for plotting. This only shifts the 
lines but does not change the slopes. 

The logarithms are then plotted giving the figure which also 
shows the true point of closing or opening of the valve as the point 
at which the straight line begins marks the beginning or ending 
of the expansion or compression. This is of value as the events, 
especially compression are difficult to fix accurately on the indi- 
cator card. 

On investigating tests on which the quality of the steam in the 
cylinder could be determined from data and on making such 
tests on an engine in the laboratory where the quality of steam 



U.8b 










— — £^0j50 








0,% 


























1.05 


















#=0.82 
37=0.86 










LI'S 


















_ #=0.90 










1.25 





50 



90 110 

Initial Steam Pressure 



150 



Fig. 94. — Clayton's relations between n, x, and p for stationary engines. 



at cut-off could be varied by using wet steam or superheated 
steam, Clayton showed that there was a variation of the value 
of n of the expansion line and the quality of steam at cut-off. 
The value of n varied from 0.70 to 1.34 while the quality at cut- 
off varied from about 0.4 to 0.9. He showed that there was 
a relation between them. In steam engines the value of n of the 
expansion line could be told from the quality at cut-off or, con- 
versely, knowing n the quality at cut-off could be found. This 
relation changes with the steam pressure and with the speed but 
does not vary with the point of cut-off. Since in stationary en- 
gines the variation with the speeds in use is slight, the only varia- 
tion considered is the variation with pressure and Fig. 94 has been 
prepared from a similar figure due to Clayton giving the values 



THE STEAM ENGINE 



213 



of x and n at different pressures. For locomotive work, where the 
pressures do not vary much, Clayton gives a figure similar to 
Fig. 95 showing the values of x and n at different speeds. 

The results have been obtained from non- jacketed engines 
exhausting near the atmospheric pressure and hence the results 
should only be applied to such. The method, if used with 
jackets or with high back pressure can only be considered an 
approximation. The indicator must be connected directly to 
the cylinder without the use of piping and a correct reducing 
motion with no stretch nor lost motion must be employed. 



0.85 




^_£^0 i 5jL 








0.95 








X^0;60_ 










^^TO^ 




8 

o 
xl.05 

a 

3 

> 








X^S£____ 










x^90__- 




1.15 








-Y— =1,00 










___x^12— 




1.25 













100 



150 200 

R.P.M. 



250 



Fig. 95. — Clayton's relations between N, n, and x for locomotive engines. 

On actually applying this method, a result close to the correct 
steam consumption may be found by assuming dry steam at 
compression. The error may be 4 per cent. The results for 
Fig. 92 will be computed. 

Value of n for expansion curve, Fig. 93, = 0.98 

Steam pressure, 108 lbs. abs. 

x, from Fig. 94, for n = 0.98 and p = 108, is 0.65 

Pressure at cut-off, 93 lbs. abs. 

Length to cut-off, 1.22 in.; to comp., 0.73 in.; of card, 3.14 in. 



Total volume at cut-off 



1.22 314X42 
X 



= 2.96 cu.ft. 



3.14 ' x 1728 

Weight of dry steam at 93 lbs. = 0.2112 lbs. per cu. ft. 

w ... - , . a 2.96X78X60X0.2112 

Weight ol steam per hour at cut-off = 



0.65 



4470 lbs. 



214 HEAT ENGINEERING 

°- 73 v. 314 X 42 1-a 
Volume at compression = ttt\ X — ^™ — = 1.76 

Pressure at compression = 24 lbs. abs. 

Weight of clearance steam per hour = 1.76 X 78 X 60 

X 0.0591 = 486 
Steam supplied = 4470 - 486 = 3984 lbs. per hour 
H.p. = 160 h.p. 

Steam per i.h.p.-hour = -t^tt = 24.8 lbs. 



VALUES OF N FOR EXPANSION LINES 

Clayton has found by logarithmic diagrams that the value of 
n for the expansion line of the steam engine varied from 0.835 
to 1.234 and that the average with saturated steam in the steam 
pipe was 0.947 while with superheated steam the value was 1.056 
and the average of all tests was 1.004. The value of n = 1 
which is so common in the theoretical discussion of indicator 
cards of steam engines is not far from the truth. The ease of 
construction of the rectangular hyperbola, pv = constant, and the 
simple results it leads to in practice are good reasons for its 
common use. Of course it has no theoretical basis and is only 
used for the reasons given above. 

In steam engines actually examined by Clayton the expansion 
line had values of n near 1 in all cases except poppet valve 
engines, where n rose to 1.3 with highly superheated steam and 
in the Stumpf Straight Flow Engine with superheated steam 
where n was 1.2. The compression lines were mostly near n = 1 
although some values much lower were found. 

In gas engines n for expansion varied from 1.09 to 1.36 while n 
for compression varied from 1.09 to 1.43. Guldner, according 
to Clayton, gives values of n from 1.30 to 1.38 for compression 
and 1.35 to 1.50 for expansion. The latter is due to hot water. 
He states that Burstall gives 1.288 as the average for expansion 
and 1.352 for compression in gas engines. 

For compressed air locomotives Clayton found n = 1.35 while 
on air compressors it was 1.26. 

For ammonia compressors the cards examined gave values 
averaging n = 1.20. . 

On gas compressors the value is n = 1.14. This is very 
low. 



THE STEAM ENGINE 215 

While discussing this value of n it is well to remember that, 
although the equation of the adiabatic of steam is 

XT 

s' + "m = const, or 



; ' + f + J^° 



dt 

const. 



the line is sometimes put in the approximate form 

pv n = const. 
n = -Q-, according to Rankine (63) 

n = 1.035 + O.lz, according to Zeuner (64) 

n = 1.059 + 0.000315p + (0.0706 + 0.000376p)z 

according to E. H. Stone (65) 

p = pressure at point where quality is x 
x = quality at highest point 

For superheated steam, 

p(v + 0.088) 1 - 305 = const. (Goodenough) 

EXPANSION LINES 

It will be well at this point to examine the difference between 
various lines of expansion which may occur in the steam engine 
cylinder to see what error might be made in assuming one rather 
than another in the theoretical discussion of indicator cards. 
The discussion and curves will also indicate how close the lines 
approach each other. 

Using the results of Clayton, suppose that an engine with cut- 
off at 34 stroke and with 10 per cent, clearance has steam of a 
quality of 0.70 at 115 lbs. absolute pressure and it is required 
to draw the following lines through this point: 

(a) Rectangular hyperbola 
(6) Adiabatic 

(c) Isodynamic 

(d) Constant steam weight, x = const. 

(e) pv n = const, of Clayton, n = 1.02 
(/) pv 1 - 2 = const. 

(g) pv°- s = const. 

From Clayton's diagram, n = 1.02. 



216 



HEAT ENGINEERING 



The diagram for 1 lb. of steam will be computed and 
tabulated below for pressures at 20-lb. intervals to points beyond 
the actual length of card. The computations for the first points 
are given. 

Vi = Mxv" = 1 X 0.70 X 3.876 = 2.71 cu. ft. 
pi = 115 lbs. 
For p 2 = 95 lbs. abs., 

(a) V 2 = 2.71 X ^f = 3.28 cu. ft. 



(b)x 2 



95 
0.4881 + 0.70 X 1.1026 - 0.4699 



1.1363 
= 0.695 
V 2 = 0.695 X 4.644 = 3.23 cu. ft. 
(b a ) Stone's value of n, 
n = 1.059 - 0.000315 X 115 + (0.0706 + 0.000376 

X 115)0.70 
n = 1.102 

V 2 = 2.71 (ijf) 1 ^ = 3.23 cu. ft. 

(c) g'i + XiPi = q' 2 + x 2 p 2 

x 2 X 808.8 = 309.0 + 0.70 X 797.0 - 294.6 

x 2 = 0.708 

V 2 = 0.708 X 4.644 = 3.28 cu. ft. 

(d) V 2 = 0.70 X 4.644 = 3.25 cu. ft. 

(e) V 2 = 2.71 (^) r5 2 = 3.28 cu. ft. 

(115\ J_ 
— ji.2 = 3.17cu. ft. 

/1 1 5\ 3 
( 9 )7 2 = 2.71 (-g5-)o.8 = 3.41cu.ft. 









Tabli 


, of Volumes 










w = l 


Adiabatic 


Isody- 
namic 


x = const. 


w = 1.02 


rc = 1.2 




Pressure 


Theory 


Stone 


n =0.8 


115 


2.71 


2.71 


2.71 


2.71 


2.71 


2.71 


2.71 


2.71 


95 


3.28 


3.23 


3.23 


3.28 


3.25 


3.28 


3.17 


3.41 


75 


4.15 


4.00 


4.01 


4.15 


4.07 


4.13 


3.88 


4.62 


55 


5.67 


5.30 


5.31 


5.66 


5.45 


5.60 


5.02 


6.82 


35 


8.90 


8.00 


8.00 


8.82 


8.32 


8.70 


7.34 


11.95 



The tables show the close agreement of the line pv 1 - 102 with the 
adiabatic and Fig. 96 shows how closely the rectangular hyper- 



THE STEAM ENGINE 



217 



bola, the Clayton line, the adiabatic, the isodynamic and the 
constant steam weight curves come together. The lines vary 
slightly. If the card abed is worked out for the various lines 
it is found that although the mean effective pressure for the lines 
pv°- s and pv 1 - 2 vary 12 per cent, from the mean of the two, the 
variation on the lines pv 1 - 02 and pv 1 - 1 which are the extremes of 
the remaining lines vary from the hyperbola by 5 per cent., or, 
5 per cent, is the maximum variation in general. The value 1.2 

120 



100 



1-1 

a 60 



8 

2 40 



20 



a 


b 










































n =1.102\ 

Adiabatic 


\*\ 


71=1^"^^ 








n = 


L20><^ 


^ Isodynair 


2 


d 








\l=1.0 















2.0 



4.0 6.0 8.0 

Volume in Cu. Ft. 



10.0 



12.0 



Fig. 96. — Variation in various expansion curves. Constant steam weight 
curve lies between adiabatic and pv 1 - 02 = const. Isodynamic can not be 
separated from curve pv = constant. 

is found on engines using superheated steam and the variation 
in the m.e.p. for the card with this line is 14 per cent, from that 
with the rectangular hyperbola. 



USE OF RECTANGULAR HYPERBOLA 

From the above it will be seen that in general the rectangu- 
lar hyperbola will give areas within 2 per cent, of the actual 
cards for ordinary engines although with superheated steam or 
very wet steam the use of the rectangular hyperbola for the ex- 



218 



HEAT ENGINEERING 



pansion line may result in an error of less than 15 per cent, in 
the area of the indicator card. Since the errors due to round- 
ing of the corners or the intersection of the various cards of a 
multiple expansion engine may lead to as great errors, the 
simplicity in the calculations and constructions when this line is 
assumed makes this curve of value in the preliminary design of 
an engine. When the probable condition of the steam is known 
the exact exponent may be used. 

CONSTRUCTION OF EXPANSION CURVES 

To draw the rectangular hyperbola the graphical construction 
shown in Fig. 97 is used. Given the original point 1, a hori- 
zontal and a vertical line are drawn through 1. If, from a, b 
and c, on the horizontal line, slanting lines are drawn to the origin 



\ 




a. 


b 


c 


a' 

/ b 


N 


I 


b" 






^-^^ 


c" 





Fig. 97. — Construction of rectangular hyperbola. 



of pressure and volume, they will cut the verticals in a', b' and c'. 
Verticals from a, b and c intersect horizontals from a', b' and c' 
in a", b" and c" which are points on the rectangular hyperbola. 

The student may see that this curve fulfills the equation of the 
hyperbola since 

Pi = Va 

v'« = V\ 

Vl = V' a 



From similar triangle ->- 

V a 

Substituting the values above 



= v n 



V' 



or 



Vl y" a 

V"aV"a = PlVl 



THE STEAM ENGINE 



219 



The graphical construction usually employed for the poly- 
tropic, pv n = constant, is open to an accumulative error and for 
that reason, although correct geometrically any slight error of 
one point is increased in the later points and hence the best way 
to construct the curve is to assume a ratio of two pressures, say 



Pi 



= r, and to use this to compute all successive points. 



Pi = rpi 
7>3 = rp 2 
Pa = rp 3 

In this way successive pressures can be read from a slide rule 
without moving the middle scales by setting the zero of the 
C scale at the value of r on the D scale. 



Since 



Pi = /vi\ n 
Pi ~ W 

Vi \r I 



The successive volumes are given by 



v 2 = r"vi 
V3 = r"v2 



The p's and v's being known, the curve may be plotted quickly. 
Thus suppose the line 

pv 1 -* = constant 

is desired to pass through p = 125 lbs. abs. and 2.5 cu. ft. Assume 



p 


125.0 


62.5 


31.7 


15.7 


7.8 


3.9 


V 


2.5 


6.6 


17.4 


46.0 


122.0 


323.0 



MEAN EFFECTIVE PRESSURE 

To find the probable mean effective pressure for an engine 
on which the cut-off and limiting pressures are known but for 
which the clearance and compression are not known as the pre- 



220 



HEAT ENGINEERING 



liminary dimensions have not been found, the best procedure is 
to assume zero clearance and zero compression and to allow for 
the effect of these by empirical constants. The card abode, 
Fig. 98, shows the theoretical card in this case. If r is the 
apparent ratio of expansion, ab will equal D/r if ed is assumed 
equal to D. The area is given by 



D D D 

area = -pi + — pi log e -jy - 

r 

area pi . ' 
m.e.p. = -p- = — [ 1 + log e r] 



pj> 



P& 



(66) 



This is the formula by which the mean height may be found if 
the line is assumed to be the rectangular hyperbola. For the 
line pv n = const. 

m.e.p. = 7 Ll+- ¥3T -J-^ 

for all values of n except 1. 

If clearance is added to the diagram as shown in the right-hand 
Fig. 98, the expansion line will be raised and for that reason the 





a T 

Fig. 98. — Theoretical indicator cards, showing effect of clearance and 

compression. 

area will be increased. On the other hand, the presence of com- 
pression in the cylinder reduces this area. The card is then 
abfcdgh. From a number of cards drawn with different clear- 
ances, pressures and cut-offs the net result of these two effects 
was to decrease the area or m.e.p. by 2 per cent. The variation 
found was from 3 per cent, to 1 per cent. 

REAL AND APPARENT RATIO OF EXPANSION 

In Fig. 98, D/on is the apparent ratio of expansion while D/om 
is the real ratio of expansion. The first is called r and it is the 



THE STEAM ENGINE 



221 



reciprocal of the cut-off. 
of expansion is 



T r 



If the clearance is ID, the real ratio 



E-+']» 



1 + lr 



The actual card ab'c'de'}', Fig. 99, differs from the theoretical 
card abcdef at several points. In most cases the steam is 
throttled or wire-drawn during admission. This causes the line 
ab in high-speed engines to drop about 10 per cent, in pressure 
although in Corliss engines the line is nearly horizontal. More- 
over, with slide valves, the slow closing of the valve causes further 
throttling and causes the corner to be rounded. The expansion 
line be is of the form pv n = const., the value of n depending on 




Fig. 99. — Actual and theoretical cards compared. 

the quality at b. The release takes place at c' before the end of 
the stroke giving the line c'd. The back pressure line should be 
that assumed, as the actual value is taken in theory. This is from 
J£ to 2 lbs. above the pressure in the region into which exhaust 
takes place. At the end of exhaust the back pressure line will 
rise due to the slow closing of the exhaust valve so that the point 
of compression e' is raised above e. This gives the line e'f 
distinct from ef. In actual cards such a point as g is considered 
as the point of compression but, as was pointed out in the loga- 
rithmic diagrams, this point is not on the compression line. 
The ratio of the actual area to the theoretical area is known as 
the diagram factor. 

Area (db'c'de'f) 

Area (abcdef) 



= diagram factor. 



222 



HEAT ENGINEERING 



The value of this for single cylinder engines is about 0.90. 
actual m.e.p. is then given by 

m.e.p. = 0.90 X 0.98[y (1 + log e r) - p 6 ] 



The 



(68) 



If the two factors are combined in a single factor known as the 
combined factor, this may be found from the results of an engine 
test by the following formula: 

actual m.e.p. 



combined diagram factor 
i.h.p. X 33000 



2FLN 



X 



theoretical m.e.p. 

1 



Pi 



(l + log e r) —p b 



(69) 



Tests give the value of this to be about 0.85 for high-speed engines. 
To find the value of the m.e.p. it is necessary to assume or 

know the values of p h p b and r. For ordinary single cylinders the 

value of pi is usually 90 lbs. 
gauge to 125 lbs. gauge un- 
less it is desired to get great 
power or force from a rela- 
tively small cylinder when a 
much higher pressure is used. 
If high pressure is used cut- 
off must be made very early 
so that the expansion of the 
steam may be utilized com- 
pletely. This, however, 
causes the percentage effect 
of initial condensation to be 




Fig. 100. — Steam consumption curves. 



felt although the higher temperature range gives a higher theo- 
retical efficiency. To reduce this loss a later cut-off is used giving 
an excessive loss due to free expansion. This same reasoning 
holds for the determination of the best value of r. If r is large 
the steam is used to advantage expansively but the percentage 
effect of initial condensation is great while for a smaller value 
of r this condensation effect is smaller but there is a loss from 
free expansion. 

If the results of a test are plotted with indicated horse-power 
as abscissae and total weight of steam per hour as ordinates, the 
most efficient point is found at the point of tangency of a straight 
line from the origin as this gives the smallest ratio of ordinate 



THE STEAM ENGINE 



223 



to abscissa which is the steam per horse-power hour. In Fig. 
100 this relation is seen for the curve AB. If the ratios of 
ordinates to abscissae are found for different points, the steam 
consumption curve CD is found on which F is the best point. 

Heck proposes that the weight of steam per hour be divided 
by the displacement per hour giving the weight of steam used per 
cubic foot of displacement. In this case the line AB would take 
the same form because the displacement per hour which is 

120FLN . , ■ . _. inn 

-, is a constant for Fig. 100. 



144 



F = area of piston in square inches. 

L = stroke in feet. 

.V = revolutions per minute. 




Mean Effective Pressure 



Fig. 101. — Curves of steam per cu. ft. of displacement for different mean 

effective pressures. 



This quantity is called m,/, weight per cubic foot of displacement, 
wt. per hr. 



Now 



m/_ 120FLN 
144 



= const. X wt. per hr. 



from which p, 



i.h.p. = 

33000 i.h.p. 
2FLN 



2pmFLN 
33000 



= const. X i.h.p. 



(70) 



In Fig. 101 the line AB drawn with p m and m f as coordinates 
is the same curve as used before with a new scale. The weight 



224 



HEAT ENGINEERING 



of steam per cubic foot of piston displacement from the indi- 
cator card, Fig. 102, when the missing quantity is m.q., is 





(I + x)D 


1 - 


1 

- m.q. 



D 



(>+t)« .+ 



v" b 



V" e 



1 — m.q. 



m/ = 



1 



(rv\)(l-m.q.) 



approximately. (71) 




Fig. 102. — Card for the computation of steam weight. 
Now the steam per indicated horse-power hour is 



Mi = 



or 



m/ 144 N X 6 ° 


13750m/ 


2p m FLN 
33000 

Mi 
m/ " 13750 


Pm 
Pm 



(72) 



(73) 



That is, straight lines from the origin of Fig. 101 are lines of 
constant steam consumption; the consumption being equal to 

771 f 

13,750 times the ratio — or the inclination of the lines. 

Pm 

The theoretical amount of steam per cubic foot of cylinder 

VOlUmeiS 1 M ° t* OTA\ 



since the volume at cut-off is 1/r and the specific weight of steam 
is 



m — a + bp 
= bp 



(75) 



if a is neglected because of small value. This results from 



THE STEAM ENGINE 



225 



plotting the specific weights and pressures of the steam tables 
as shown in Fig. 103. This shows clearly that the value of m 
is practically bp. 



Now 



Pr 



kM x = 



Pi 



[1 -f loge r] - p h 



k 



Pi 



rtif 


r 




Vm 


&[l + log.r] 
1 


- Pb 



[1 + loge r] — kr 



Pb 
Pi 



(76) 



In other words Mi decreases as the expansion increases since 



log e r increases faster than r 



Pb 
Pi 



lit) 

150 

a 

& 

" 125 

o 

3 


















































.2 100 

a 

3 

1 75 

S 

03 


































a 50 

9 
Si 


















^ 25 



















0.05 



0.10 0.15 0.20 0.25 0.30 0.3 

Weight of 1 Cu. Ft. of Saturated Steam in Lbs. 



0.40 



Fig. 103. — Curve of relation between pressure and weight of 1 cu. ft. of 

steam. 



The amount of initial condensation per cubic foot of cylinder 
displacement varies slightly with r; that is, as p m increases 
the condensation increases slightly. CD will represent the 
initial condensation in Fig. 101. The distance from AB to DC 
will be the amount shown by equation (74). From this figure 
Mi can be found as well as from (76). 

15 



226 HEAT ENGINEERING 

The best point is usually found at about J£ or % cut-off with 
single cylinders of the ordinary form. 

While discussing steam consumption it will be well to call 
attention to the Willans straight line law for the steam con- 
sumption of throttle-governed engines. For a given ratio of 
expansion and back pressure 

Pm = Cpi — Pb 

_ 2(cpi - p b )FLN 
l.n.p. 3300Q 

_ 2LFN X 60 kpi _ 2LFN 60fc r 33000 i.h.p. 1 

M total - 144 X r - 144 X cr I 2Fm + p b \ 

= ¥ i.h.p. + h"p h = a i.h.p. + b (77) 

or, the total weight of steam per hour with a throttle valve 
engine varies with the horse-power on a straight line. 

STUMPF ENGINE 

Although in the ordinary engine the best point of cut-off seems 
to be about one-fourth stroke, in a single cylinder engine developed 
by Prof. Stumpf in 1910 the cut-off is carried very early since 
the initial condensation is reduced by the arrangement of the 
engine. This is the uniflow (unidirectional flow) or straight 
flow engine shown in Fig. 104. This was invented independently 
by Stumpf although a similar idea had been developed by 
Bowen Eaton in 1857, by E. Roberts in 1874 and by L. J. Todd 
in 1885. In this engine steam is allowed to enter from the 
valve A behind a long piston B and forces it to the right. After 
the piston moves a short distance to the right, steam is cut off by 
the double beat valve and the steam expands to nine-tenth of the 
stroke when the piston overrides ports C, cut in the cylinder barrel, 
allowing steam to exhaust through D to the condenser or atmos- 
phere. The area thus available is so great that the pressure 
rapidly falls and when the piston returns and covers this the 
steam has dropped to the pressure of the condenser or atmosphere. 
The compression then begins and, by properly selecting the 
clearance, the final pressure is made equal or nearly equal to the 
boiler pressure. This clearance depends on the back pressure 
and, if this is liable to change very much, arrangements are made 
to vary the compression or clearance. Thus, with a condensing 



THE STEAM ENGINE 



227 



cylinder, when it is necessary to start at atmospheric pressure 
until the air pump connected with the engine has produced the 
proper vacuum, an auxiliary clearance volume is connected to 
the cylinder and is controlled by a valve so that it may be con- 
nected when necessary. 

As the steam expands in the cylinder, it becomes wet due to 
the work done, as is known from the discussions of the adiabatic 
for steam. The supply steam in the hollow head D, however, 




Fig. 104. — Cylinder of Stumpf engine with indicator card. 



warms the steam in the cylinder in contact with the head and 
keeps it dry or superheats it as the pressure falls. As the steam 
is discharged outward at the end of the stroke the steam next 
to the head expands, driving the wet steam before it, and leaves 
the cylinder practically full of dry steam. As the piston returns 
the steam compressed is superheated as it starts from a dry or 
even superheated condition. This steam gives up heat to the 
piston head so that the steam entering the cylinder meets a 
piston surface that is warm while the head and walls are heated 



228 



HEAT ENGINEERING 



to the temperature of the incoming steam on one side and to a 
temperature higher than this by the superheated steam on the 
other side. The part of the cylinder near the center which is in 
contact with the low pres- 

sure or exhaust steam is 
cut off by the piston from 
contact with the fresh steam. 
The curves of Fig. 105 made 
by Stumpf show how the 
wall temperature (a) varies 
along the cylinder of a con- 
densing engine compared 
with the temperature (b) of 
the saturated steam corre- 
sponding to the pressure 
when the piston is at various 
points. The fact that the 
metal line is above the satu- 
ration line indicates that 
superheated steam must be 
present and this is only possible on compression. The cool 
central portion of the cylinder is a good feature structurally 
for lubrication as at this location the piston is moving at its 




Fig. 105. — Curve showing temperature 
of wall and saturated steam for different 
positions of piston as found by Stumpf. 




Mean Effective Pressure 



Fig. 106. — Steam consumption and heat per h.p. hr. curves of Stumpf 

engine. 

highest speed. Stumpf claims that the steam in this engine 
does not enter in a disturbed condition and that it flows 
always toward the exhaust. This, of course, cannot be true for, 



THE STEAM ENGINE 229 

until release, the action is the same as in any engine, but, at 
release, the steam which has been dried or superheated by the 
jacketed head drives out the wet steam, whereas, in the ordi- 
nary return flow engine, this steam is the first to leave. The 
distribution of the exhaust around the circumference gives an 
undisturbed discharge although it tends to remove any moisture 
from the piston face. The jacket head being in the steam supply 
is a bad feature as this would introduce wet steam into the 
cylinder unless superheated steam is used. The effect then 
would be to reduce the amount of superheat only and in one test 
this amounted to 54° F. or 27 B.t.u. per pound of steam. The 
chambers E and F are jackets for the barrel and the steam 
gradually passes from one to the other and with this a drop in 
pressure causes a drop in temperature. To show the efficiency 
of this engine, Fig. 106 is presented. This is from a test of a 
300 h.p. engine tested with saturated steam and superheated 
steam of 580° F. at 135 lbs. absolute pressure. The results are 
plotted for pounds per horse-power hour and actual B.t.u. per 
horse-power hour chargeable to the engine plotted against 
m.e.p. They show the closeness of results with saturated and 
superheated steam, the slight variation with great change in load 
and values, which are ordinarily obtained with multiple expan- 
sion engines. This engine illustrates the great value of expansion 
when the effects of initial condensation can be eliminated. 

SIZE OF ENGINES 

To find the probable size of an engine to deliver a given horse- 
power, the formula for horse-power is used 

b.h.p. _ 2pFLN 
mech. eff. 33000 u ; 

p = m.e.p. in pounds per square inch. 

F = area of piston in square inches. 

L = stroke in feet. 

N = revolutions per minute. 

In this equation p has been determined from the formula 
(68) after p h p b and r have been assumed. N is fixed by the 
purpose for which the engine is to be used or by the valve gear. 
For Corliss engines, N varies from 60 to 120. For connected 
valve gears, N may be as high as 450. The allowable piston 



230 HEAT ENGINEERING 

speed, 2LN, then fixes L. This varies from 250 in small engines 
to 1000 in larger engines. The only unknown is now F by which 
the diameter is fixed. 

To apply this, suppose it is desired to find the size of an engine 
to drive a 125 kw. generator. 

kw 
Lh - P - = 0.746 X0.90 X o.90 = L6kw - = L6X125 = 200ih -P- < 79 > 

This number, 1.6, is an important average number to keep in 
mind for rapid computations. The following assumptions will 
be made: pi = 125 lbs. abs., p b = 17 lbs. abs., r^= 4, 2LN = 
500, N = 275. 

m.e.p. = 0.90 X 0.98 [^ (1 + 2.3 X 0.602) - 17] 

= 50.9 lbs. per square inch 

L = ^ = 0.91 ft. = 10.9 in. 

i.h.p. X 33000 _ 200 X 33000 
b ~ 2pLN " " ~ 2 X 50.9 X 0.91 X 275 " 59 Sq ' m " 



..g.^pm.^ 



The size of cylinder necessary would therefore be 18 in. di- 
ameter X 12-in. stroke, the somewhat unusual ratio of diameter 
to stroke being caused by a desire to keep the piston speed low. 

The effect of clearance on an engine is largely dependent on 
the amount of surface exposed. The volumetric clearance does 
not seem to have much effect on the efficiency. Tests have 
been made on an engine with varying amounts of clearance and 
the results have shown little change in steam consumption. 
The effect of clearance surface as shown by Heinrich has been 
mentioned on p. 209. General practice, however, is to cut the 
clearance to a minimum so as to increase the area of the indicator 
card for a given displacement and to make the steam loss due to 
improper compression with different cut-offs as small as possible. 

BEST POINT OF COMPRESSION 

The effect of compression is very slight as is shown theoretically 
and by the tests of Heinrich but to determine the best point 



THE STEAM ENGINE 



231 



theoretically several methods may be used. The amount of 
compression to be used with any given cut-off may be found by 
the following construction of Stumpf. In Fig. 107 the initial 
pressure is pi, and the back pressure p b , the curves are rectangular 
hyperbolae. With the dimensions given on the card, 

Area = pi f- PiD [I + -] log e z — Dp b [l — x] 



+ 1 



p b D[l + X] l0g e 



l + X 

I 



It is desired to eliminate x in terms of the volume of steam taken 
from the boiler, that is, x"D. 



xD 




Fig. 107. — Stumpf s method of finding compression point. 
Hence Dp b (l - x) = Dp b [l + I] - Dp b [l + x] = 



DpJLl + 1]- D Vl [l + 



1 



x"] 



and 



p b D(l + x) log l -±* = Pl D(l+ }~x") log e A_!L^ 1^ 



m.e.p 



area _ p\ 
~D = 7 



+ Pill + 



+ p,[z+±]log.^-pJl + Z] 



~ + l 

r 



Vl [l+--x"] log. 



[t + h*'[ 



Vi 



Suppose now that the volume, x"D, of working steam re- 
mains constant and there is a change in the point of cut-off. 



232 



HEAT ENGINEERING 



The relation which gives maximum work under this change with 
constant volume of working steam is given by equating the 
derivative to zero. 

rf(m.e.p.) n , . l+l [ 7 ^l y +l l + l 

1 + --x" 



+ Vi - Vi log* 



2i 



Pid + 



l 



x") 



1 



1+-- x" 
r 



Pi log, 



l + l 

X - + l 

r 

l + l 

'-+1 



I + 



Pi log, 



l+x 
1 



Pi , l + x 

- = Pi log, - r 



(80) 



or the point of cut-off divides a line from the end of the card to 
the axis of volume in the same proportion as the clearance di- 
vides a line to the axis of volume from the point of compression. 



D A 






* v 




^h a" 






H' 





Fig. 108. — Webb's method of finding compression point. 



Another method of finding the point of compression as given 
by Webb in the American Machinist for 1890 is to carry out 
the expansion line to the back pressure line as in Fig. 108 and 
then lay off CB so that the area ABCD is equal to area EFG. 
The point C is then taken as the end of compression. The basis 
of this construction is the fact that, by changing the area 
ABCD to EFG, the card CBEFGH will have the same area as 
ABEGHC and will use the same weight of steam since this is 
proportional to the line CB. The card CBEFGH has complete 



THE STEAM ENGINE 233 

expansion and must have complete compression to overcome the 
effect of clearance. For instance, if the line H'C were used for 
the compression line, the area C'CHH' would have to be saved 
to make up for the additional steam CC since 

C'CHH' _ CBEFGH 
CC " CB 

But the area JC'C is not obtained and hence the gain of area 
JCHH' is not proportional to the increase in the steam quan- 
tity CC. 

If the line were H"C"K, then the saving in steam would be 
CC" but the lost work would be H"HCKC". This is greater 
than the proportional reduction H"HCC" by the area CKC" 
and hence this does not pay. 

Figs. 107 and 108 are for the same conditions and the results 
are practically the same. 

Of course the constructions above are true if the steam is as- 
sumed dry or proportional in all cases to the length of the line 
between the compression and expansion lines. 

The methods of Clark, see his Steam Engine, p. 399, and of 
Ball, A.S.M.E., xiv., p. 1067, may be referred to by the student. 
Clark's method is the equivalent of the two methods given 
above although the results are given in the form of a table. 

Having the compression desired, the pressures, the cut-off, 
release and clearance from the design of the cylinder, the probable 
card for an engine may be drawn and studied. 

SPEED 

The speed of an engine affects its efficiency in changing the 
amount of condensation. As this varies inversely as the cube 
root of the number of revolutions per minute it would naturally 
be expected that high-speed engines would be the more efficient. 
That this is not so is due to the fact that the term s varies 
inversely as the linear dimension of the engine and hence this 
term is so small for large engines that it overbalances the effect 
of the slower speed. For instance, if the piston speed, 2LN, is 
fixed, the initial condensation will vary approximately inversely 
as the cube root of N and also inversely as the square root of 
L. Hence it would pay to make L large and N small. This is 
the actual result in practice. The large, slow-speed pumping 



234 



HEAT ENGINEERING 



engines represent the best type of engines using saturated 
steam. 

The effect of superheat has been explained and results given 
at the end of Chapter II show that this materially affects the 




condensation and increases the efficiency a greater amount than 
that estimated by theory. 

The effect of the jacket on a cylinder is a debated question. 
The results shown by the committee of the Institution of Me- 



THE STEAM ENGINE 235 

chanical Engineers of Great Britain indicate a distinct gain. 
Other tests made on large engines show no gain and, in a few 
cases, a loss. Of course the total steam used by the engine is 
always considered in these cases. In most instances the important 
saving has been found on small engines of less than 300 h.p. 
For large engines the results will vary. Ten to 15 per cent, 
may be saved on engines of 200 or 300 h.p. One form of jacket 
which has been used on the type of semiportable engine known 
as the locomobile is of value. In this unit the engine is mounted 
on top of the boiler, Fig. 109, and the exhaust gases from the 
boiler pass around the cylinder, thus heating it to a high tempera- 
ture. These units have the condensation reduced to such a de- 
gree by this jacket combined with the use of superheated steam 
that they give a brake horse-power hour on 10^ lbs. of steam 
1.25 lbs. of coal or 210 B.t.u. 

Effect of regulating by throttling steam and changing cut-off 
may be seen when the formula (53) is noted. In this the 
quantity s changes slightly with the cut-off and so there is a slight 
increase in the steam due to this. With throttling, however, 
there is an amount of available energy lost apparently. When it 
is remembered that this energy is used in changing the quality 
of the steam which becomes drier or superheated, it is seen that 
when the steam enters the cylinder it is in such a condition 
that the initial condensation is less and hence the loss due to 
this effect is smaller. So great is this reheating effect with 
throttling that, even with this steam of reduced availability the 
steam consumption is as low as with engines in which the cut-off 
is varied. A number of years ago the author experimented on a 
small engine, controlling it by automatic cut-off and by throttle 
governing, and both curves, similar to CD, Fig. 100, were coin- 
cident throughout their range. The curves will indicate the 
steam for either of these methods and also how the consumption 
varies with the load. 

TOPICS 

Topic 1. — Sketch the Rankine cycle of the steam engine with complete 
expansion and compression and explain the difference between this and the 
Carnot cycle. Explain the variation in quality on the various lines in these 
two cycles. Show that clearance has no effect in this case when initial 
condensation is not present. 

Topic 2. — Sketch the Rankine or Clausius cycles with complete expansion 
and incomplete expansion on the y-v and T-S planes and derive the various 



236 HEAT ENGINEERING 

expressions for efficiency. Explain why the cards may be drawn with no 
compression lines. 

Topic 3. — Sketch the T-s diagram of a Rankine cycle with incomplete 
expansion and discuss the value of increasing the steam pressure, vacuum or 
amount of superheat. Does the increase of vacuum always pay? Why? 

Topic 4. — Sketch and explain action of the Barrus calorimeter. Give 
the method of computing results. Explain what is meant by a dry test, how 
it is made and for what it is used ? What is a separating calorimeter ? What 
is the formula for x when this instrument is used. What is an electric calo- 
rimeter? What are constant immersion thermometers? Why are they 
Used? What is the purpose of the upper thermometer of the Barrus 
calorimeter? 

Topic 5. — What is Hirn's analysis? For what is it used? What observa- 
tions are made? Explain how the quantities M and M are found. Ex- 
plain how Xi, x 2 , and x& are found. Explain how AUi, AU2, AU3 and AUi 
are found. Explain how Qi and Q 2 are found. 

Topic 6. — Given the average indicator card, show how the quantities 
AW a , AWb, AW C and AW a are found. Give method of changing from 
square inches to B.t.u. Having the U's, Q's and AJF's explain how Q a , Qb, 
Q c and Qa are found. What check equation may be used for these quanti- 
ties ? How is the new quantity for this check equation found ? What differ- 
ence does a jacket make in these equations? 

Topic 7. — What is the temperature-entropy analysis? What readings are 
necessary? Explain how the curves of the quadrants are constructed and 
sketch the diagram. How is the average indicator card found and how is 
this transferred to the p-v quadrant of the diagram. 

Topic 8. — Explain how to transfer the diagram from the p-v quadrant to 
the T-s quadrant. Explain the method of finding the various losses. Which 
lines on the diagram are true? Why can the diagram be used for other 
lines? 

Topic 9. — What is initial condensation? How large may it be? To what 
is this due? What is the missing quantity? In what units is it found? 
Explain how it may be found from a test. Explain what is meant by the 
terms of the two f ormulge below : 



mq 



30 VT 
1 - rnq d VN 

mq 



027 uf 
"Z/N \ pe 



Topic 10. — Derive the formulae: 



M, =K'Fc^fT 



Why is it that it may be said that v"i = ? What is Heck's expression 

0.27 jsf 
for s? Using Heck's formula m.q. = 3/ ^ \j discuss the effect of size, 



THE STEAM ENGINE 237 

speed, pressure and cut-off on the actual condensation and on the percent- 
age condensation, m.q. 

Topic 11. — Give a statement of the work of Callendar and Nicolson, 
Hall, Duchesne and Adams on the effect of cylinder walls. How do Callendar 
and Nicolson plot their results so that initial condensation may be found? 
What do their results show in regard to variation of steam temperature in the 
cylinder, and in a small hole in the head ? What is the nature of the tempera- 
ture cycle at various points in the cylinder length and at various depths? 
Which has the greater effect, clearance surface or clearance volume? When 
does the major part of the initial condensation take place? To what is it 
due? 

Topic 12. — Explain Clayton's method of finding the quality at cut-off, 
constructing the diagram from the indicator card. Show how to find the 
probable steam consumption from the indicator card. Explain how leaks 
are detected. Explain how to locate the events of the stroke properly. 

Topic 13. — What values of n are to be expected on various machines on 
compression and expansion? Explain how the following curves may be 
constructed on the pv-plane: (a) rectangular hyperbola, (6) adiabatic, (c) 
constant steam weight curve and (d) pv 1 - 05 = const. What conclusions 
may be drawn from the figure in the book showing the various expansion 
curves? 

Topic 14. — Explain how to construct the curves pV = const, and pV n = 
const, through a given point piVi. Find the area of the indicator card for 
each of these to a volume V% and pressure p 2 if the back pressure is pi. From 
this find the formulae for the m.e.p. 

Topic 15. — Explain what is meant by the real and apparent ratio of expan- 
sion. Derive a formula for the real ratio of expansion in terms of the 
apparent ratio, r, and the clearance, I. What is meant by diagram factor? 
How is it found? What value does it have? 

Topic 16. — Sketch the curve of total steam consumption and show how to 
find from this the curve of steam consumption per horse-power hour. Ex- 
plain why this curve is the same as the curve between m/ and p m . What is 
the Willans straight-line law? Prove that it is true. 

Topic 17. — What is the Stumpf engine? Why is it of value? Sketch it. 
Sketch curves showing the temperatures of the wall and saturated steam for 
various positions along the length of the cylinder. What conclusion may 
be drawn from this? Sketch the cards from this engine. How is compres- 
sion cared for when high back pressure exists in starting a condensing Stumpf 
engine with a direct-connected air pump? 

Topic 18. — Explain how to find the indicated horse-power required to 
produce a given kilowatt output from a direct-connected generator. Show 
how to find the size of an engine to produce this power. 

Topic 19. — Derive the formula for the best point of compression. 

Topic 20. — Sketch the constructions for the best point of compression due 
to Stumpf and to Webb. What is the effect of speed? Of superheat? Of 
jackets? What difference is found between governing by throttling and by 
automatic cut-off? 

Topic 21. — Sketch and explain particular features of the locomobile. 



238 HEAT ENGINEERING 



PROBLEMS 



Problem 1. — Find the various efficiencies of an engine with p x = 120 lbs. 
gauge pressure, p 2 = 30 lbs. gauge pressure, p 1 = 2 lbs. gauge pressure if xi 
= 0.99 and 35 lbs. of steam are needed per i.h.p.-hr. 

Problem 2. — One engine uses 25 lbs. of steam per i.h.p.-hr. with steam at 
125 lbs. gauge pressure and of quality 0.995. The back pressure is 2 lbs. 
by gauge. By using steam of 200° F. superheat the steam consumption is 
reduced to 22 lbs. per i.h.p.-hr. What was the saving? 

Problem 3. — An engine uses 3500 lbs. of steam per hour, of which 300 lbs. 
is used in the jackets and 200 lbs. in the receiver. The steam supply is at 
150 lbs. gauge pressure with 175° F. superheat. The temperature of the 
return from the jackets is 320° F., while the return from the receiver is 338° 
F. and the hot well temperature from the condensate is 95° F. The engine 
develops 250 h.p. Find the B.t.u. per h.p.-min. Find the actual efficiency. 
If the mechanical efficiency of the pump and engine combined is 92 per cent., 
what is the duty of this engine? 

Problem 4. — In Fig. 68 assume that the pressure on the top line is 120 
lbs. absolute and on the lower line is 15 lbs. absolute. Suppose that the qual- 
ity on the top line is 0.98, assuming the cycle to be the Rankine cycle, and 
that it varies from 0.03 to 0.98 if assumed to be the Carnot cycle. Find the 
qualities at the lower corners. Find the heats on the four lines. 

Problem 5. — In a Rankine cycle with complete expansion the pressure 
varies from 125 lbs. gauge to lbs. gauge with x\ = 1.0. Find the efficiencies, 
771, 772, 773- Increase the upper pressure to 150 lbs. gauge and, leaving the other 
quantities unchanged, find 771, 772 and 773. With 125 lbs. initial gauge pres- 
sure assume the quality changes to 160° F. superheat; find 771, 772 and 773. 
Assume the back pressure is changed to a vacuum of 27 in. but with no 
change in other conditions; find 771, 772 and 773. 

Problem 6. — In a Rankine cycle with incomplete expansion let the initial 
gauge pressure be 125 lbs., the pressure at the end of expansion 20 lbs. gauge 
and the back pressure is that of the atmosphere. If Xi = 1.0 find the three 
efficiencies, 771, 772 and 773. Change the initial pressure only to 150 lbs. gauge 
and find the efficiencies. Change the initial quality to 160° F. superheat 
and find the efficiencies. Change the back pressure only to 27 in. and 
find the efficiencies. 

Problem 7. — The following results were obtained from a test of an engine : 

Size of engine 10 in. X 14 in. (neglect rod) 

Time of test 60 min. 

Clearance 7 per cent. 

Number of revolutions 15,000 

Steam used 3003 lbs. 

Average gauge pressure at throttle 112 lbs. per sq. in. 

Barometric pressure 14.7 lbs. per sq. in. 

Average quality of steam 0.99 

Average temperature of condensate 135° F. 

Average temperature of water leaving 120° F. 

Average temperature of water entering 75° F. 

Weight of condensing water 58,100 lbs. 



THE STEAM ENGINE 239 

Average results from indicator cards 

Point of admission 0.0% of stroke, .... 55 lbs. abs. per sq. in. 

Point of cut-off 33 . 0% of stroke, .... 114 . 7 lbs. abs. per sq. in. 

Point of release 93 . 0% of stroke, .... 44 . 7 lbs. abs. per sq. in. 

Point of compression 20.0% of stroke,. ... 14.7 lbs. abs. per sq. in. 

Work of admission 3.22 sq. in. 

Work of expansion 3.32 sq. in. 

Work of exhaust — 0.70 sq. in. 

W^ork of compression — 0.46 sq. in. 

Make Hirn's analysis from the above data. 

Problem 8. — In the analysis above the following coordinates give the 
positions of the points in inches on a 4-in. card from the true zero of pressure 
and volume with a spring scale of 50 lbs. per inch. 

Points Volume Pressure Points Volume Pressure 

1 0.28 2.48 10 4.28 0.29 

2 0.78 2.44 11 3.28 0.29 

3 1.28 2.36 12 2.28 0.29 

4 1.60 2.28 13 1.08 0.29 

5 2.28 1.60 14 0.78 0.41 

6 3.28 1.19 15 0.48 0.63 

7 3.78 0.94 16 0.28 1.08 

8 4.02 0.88 17 0.28 1.78 

9 4.18 0.60 18 0.28 2.30 

Plot the card and make the T-S analysis. Find the various losses. 

Problem 9. — Find the probable value of the missing quantity for the engine 
given in Problem 8 by the various formulae of this chapter. 

Problem 10. — If the ratio of connecting rod to crank is 6 to 1 mark the 
piston positions for every five-sixtieths of a revolution on the indicator card 
constructed from the data of Problems 7 and 8. Also find crank positions of 
events of stroke. Construct the Callendar-Nicolson diagram for this card 
and find the probable initial condensation. Express this as a percentage of 
the steam given in Problem 7 and compare results with those of Problems 
7, 8 and 9. 

Problem 11. — Using data of Problem 8 construct table and diagram used in 
Clayton's method. Find probable initial condensation and steam consump- 
tion. Compare this with other results. 

Problem 12. — Find the real and apparent ratios of expansion of the card of 
problems 7 and 8. Find the diagram factor. Find the horse-power de- 
veloped in Problem 7. Find the actual steam consumption. 

Problem 13. — Construct a curve of total steam consumption and from it 
find the curve of steam per i.h.p.-hr. 

Problem 14. — Find the size of a single-cylinder non-condensing engine to 
drive a 125-kw. generator with initial gauge pressure of 130 lbs., cut-off at 
0.3 stroke and a back pressure of 2 lbs. gauge. 

Problem 15. — Construct the card for Problem 14 and find the best point 
of compression for 10 per cent, clearance by Stumpf's method and Webb's 
method. Release is at 95 per cent, of stroke. 



CHAPTER VI 



MULTIPLE EXPANSION ENGINES 



Multiple expansion, which is the use of steam in one cylinder 
after another, was introduced to cut down the wastes in steam 
engines. After its introduction it was seen that for structural 
reasons such an arrangement is of value, as it reduces the sizes 
of parts of the machine and gives a more uniform turning 
moment. The reason for the reduction in waste is the fact 
that there is a series of small ranges in temperature; that 

some of these ranges act on 
small surfaces (i.e., in the 
smaller high-pressure cylin- 
ders), and lastly that heat ab- 
stracted by the exhaust steam 
from the upper stages may 
be of value for use in the lower 
stages. If the indicator cards 
from a two stage or compound 
engine are taken as shown in 
Fig. 110, these cards may be 
used as shown in the preced- 
ing chapter to study the ac- 
tion of the cylinder walls, to 
find the horse-power and to 
use for any purpose that cards 
for simple engines are used. However, to get a fuller picture 
it is often desired to construct what is known as a combined dia- 
gram. To form this the cards of Fig. 110 from a 10 and 20 X 24 
engine are divided by ordinates at regular intervals, say 10, after 
laying off the clearance at the end of each diagram. The extreme 
ends of the diagrams are placed on a new axis of volume and the 
base lines with their ordinates are drawn in after making the card 
lengths proportional to the volumes swept out by the respective 
pistons. In most cases since the strokes are equal these are pro- 
portional to the squares of the diameters. This increases or de- 

240 




Fig. 110. — Indicator cards' from high 
and low pressure cylinders of com- 
pound engine. 



MULTIPLE EXPANSION ENGINES 



241 



1 

'I 

{ , 


30 ^ Spring Scale 








^> 
















^ 












-— - 


"^ 

















creases all horizontal dimensions. If now a scale be assumed for 
pressure and the ordinates from the atmospheric pressure line are 
measured on the cards and reduced to this new scale, the points 

may be plotted on the new 
figure and the cards of Fig. 
Ill drawn. These are both 
drawn to the same scales of 
volume and of pressure. The 
diagram is known as the com- 
bined diagram. Since, how- 
ever, the weights of clearance 
steam in each cylinder are not 
the same, although the work- 
ing steam is equal on each, 
these figures do not represent 
diagrams for the same total 
weight of steam on the ex- 
pansion line. For this reason 
Fig. 111. — Combined cards for com- , u ,. „ . 

pound engine. tne expansion line of one cyl- 

inder does not pass through 
that of the other. The same is true of the compression lines. 
These cards show in this figure the relative amounts of work 
done by each cylinder and the amount lost due to the drop in 
pressure between the two cylinders. That these diagrams cut 
each other means noth- 
ing since the diagrams 
are not simultaneous for 
the same piston positions. 
The first cylinder is 
known as the high -pres- 
sure cylinder and the 
second as the low-pres- 
sure cylinder. If there 
are three stages, known 
as triple expansion, the 
cylinders are known as I ^-^ 

high pressure, interme- p IG 112.- 
diate pressure and low 
pressure. In addition 

to high- and low- pressure cylinders there are first and second in- 
termediate pressure cylinders in the quadruple expansion engine. 

16 



6 1 








5 \ 


\ 2 






13^4 


J 3 


-L8 




1 12 










11 




^ -1 









-Combined cards from Woolf com- 
pound engine. 



242 HEAT ENGINEERING 

When the cylinders and cranks are so arranged that when the 
high-pressure cylinder is ready to discharge steam the low-pres- 
sure cylinder is ready to receive it, the engine is known as a 
Woolf compound engine, while if the discharge must be stored 
in a receiver before it can be used, the arrangement is known as 
a receiver engine. At times several cylinders may be used for 
any stage, say low, for the purpose of reducing the size. For the 
diagram of the Woolf engine see Fig. 1 12. . The cut-off in the high- 
pressure cylinder occurs at 1 and the steam expands to 2. At 
this instant the steam is connected to the low-pressure cylinder 
in which the pressure is that shown by 12 together with the con- 
necting pipe in which the pressure was left the same as at 8. 
Hence the pressure falls from 2 to 3 in the high and rises from 12 
to 13 in the low and from that of 8 to that of 3 or 13 in the 

8 1 



Fig. 113. — Combined cards from receiver engine. 

receiver or connecting pipe. The pressure at 13 is the same as 
that at 3 except for a drop due to throttling in the passage through 
the connecting pipe. As the high pressure forces the steam out 
it enters the low-pressure cylinder in which there is a larger 
piston moving at the same speed as that of the high-pressure 
cylinder, and the pressure drops gradually due to this increase of 
volume. This finally reaches the point of compression on the 
high-pressure cylinder which corresponds to the point of cut-off 
on the low-pressure cylinder in many cases. The pressure 
at 7 and that at 4 are about the same. The line from 4 to 5 is 
the compression and from 5 to 1 there is admission. The steam 
in the low-pressure cylinder and connecting pipe or receiver now 



MULTIPLE EXPANSION ENGINES 243 

expands from 7 to 8 at which point the cut-off occurs in the low- 
pressure cylinder. From 8 to 12 the events on the low are the 
same as those on a single cylinder. 

The combined cards from a receiver engine are shown in Fig. 
113. In this case at the end of expansion the high-pressure 
cylinder discharges into the receiver in which the pressure is that 
at 11 if cut-off in the low pressure occurs after compression on 
the high-pressure cylinder, while the pressure is that of 6 if this 
compression in the high occurs later than the cut-off in the low. 
In either case, the pressure in the receiver is usually lower than 
that at 2 and there is a drop in pressure. The low-pressure 
cylinder is not in a position to take steam at this point as in most 
cases the cranks of the engines are at right angles and for that 
reason the receiver is used. The amount of drop 2-3 not only 
depends on the pressure in the receiver but also on the relative 
volume of the receiver and the cylinder. This may be lj^ times 
the volume of the high-pressure cylinder. As the piston now 
returns the steam is compressed into the clearance space and the 
receiver to the point 4. Here steam enters the low-pressure 
cylinder in which the pressure is 15. There is a drop from 4 to 
5 in the receiver and high-pressure cylinder and a rise from 15 to 
9 in the low. The line 3-4 is a rectangular hyperbola with the 
origin at a distance of the receiver volume from the clearance 
line. From 5 to 6 the small piston at the middle of its stroke is 
moving so much faster than the large piston at the beginning of 
its stroke that the volume decreases and the pressure rises at 
first. It then usually falls before 6 is reached, 9 agrees with 5, 
and 10 with 6. Cut-off on the low occurs after compression on 
the high but before the high pressure reaches the end of its stroke. 
If it did not occur sooner than this there would be a second 
admission of steam from the other end of the high-pressure 
cylinder. The other points are clearly seen. 

COMPUTATION OF CARDS FOR CONSTRUCTION 

To compute these various points the volumes of the two 
cylinders must be known, with the clearances; the volume of the 
receiver must be known, and finally the points of all the events 
of each card and the initial pressures. 
Let V r = receiver volume. 

V = volume of cylinder at event represented by sub- 
script including clearance. 
p with subscript = pressure at point. 



244 HEAT ENGINEERING 

Now from the curve on Fig. 103 it is seen that m = kp ap- 
proximately, hence the total weight of saturated steam at any 
point is 

M = mV = kpV (1) 

or the weight of steam is approximately equal to the product 
pV. Hence if quantities of steam of different volumes and at 
different pressures are connected the resultant pressure mul- 
tiplied by the sum of the volumes must be equal to the sum of 
the individual products of pressure and volume, or 

2pV 



With this understanding, the following equations hold: 
Pi7i 



(2) 



p% = -^ (known) (3) 

V 2 

P* = y _f y \P 11 unknown) (4) 

Pi = — Y i _j_ T/ ( In terms of pu) (5) 

_ PiiVj + Vr) + P15F15 dn terms of p n ; since . . 

P5-P9- F4+Fr+Fi5 puVlb i s known) W 

P15F15 = P14F14 (7) 

P 5 (F 4 + Vr + F 15 ) „ . . ' , Q . 

p & = — y g + y^ + y (In terms of pu) (8) 

P&Vq . . 

Vi = ~YV (9) 

Pw(Vio-{-Vr) n . . f . • , \ nt\\ 

pu = ^7 — : 17 . (In terms 01 pu; . . pu is known) (10) 

HIT Vr 

PnVn n .s 

Pl2=~r^- (11) 

K12 

To apply these a table is first computed for all volumes in pro- 
portional numbers if not in actual numbers, and from this table 
the substitution can be made in the equations above. To find 
the relative position of events a diagram is made with the cranks 
as shown in Fig. 114. The actual position could be found by 
using the connecting rods or if infinite rods are assumed the pro- 
jections may be used. Thus if the compression is to occur at 
0.1 stroke on the high-pressure cylinder, the diagram for an in- 
finite connecting rod would be shown in Fig. 114, with a crank 



MULTIPLE EXPANSION ENGINES 245 

of 0.5. From this it is seen that the low-pressure piston will 
have moved 0.2 of its stroke when compression occurs in the 
high-pressure cylinder. The high-pressure piston is at the middle 
of the stroke when the low-pressure stroke begins. 

Suppose that the cylinders have a ratio of 1 to 4 and the 
clearance is 6 per cent, on the high and 5 per cent, on the low. 
Suppose that the receiver volume is 0.8 times the volume of the 
high-pressure cylinder; that the cut-off occurs at 0.33 stroke in 
high-pressure cylinder and 0.45 stroke in low; that compression 
is at 0.1 stroke from end on high and 0.2 on low and that pi 
is 150 lbs. gauge and pb is 2 lbs. absolute. 




Fig. 114. — Diagram by which to find piston position. (Finite or infinite 

connecting rod.) 

Call the volume swept out by the high-pressure cylinder 1 and 
using Fig. 113: 

V h = 1 Vi = 0.33 + 0.06 = 0.39 

Vi = 4 V 2 = 1.00 + 0.06 = 1.06 = V 3 

V r = 0.8 7 4 = 0.50 + 0.06 = 0.56 = V b 

V ch = 0.06 X 1 = 0.06 7 6 = 0.10 + 0.06 = 0.16 

V cl = 0.05 X 4 = 0.20 V 7 = 0.06 = V 8 

V 9 = 0.20 = Fib. 
Fio = 0.20 X 4 + 0.20 = 1.00 
7u = 0.45 X 4 + 0.20 = 2.00 
7 18 = 4 + 0.20 = 4.20 = V 13 
V u = 4 X 0.20 + 0.20 = 1.00 

p 2 = 164.7 X ™| = 60.6 
1.06 

60.6 X 1.06 +pn X 0.8 Q . K . A , Q 
V > = 1.06 + 0.8 = 34 ' 5 + °- 43 ^ 11 

(34.5 + 0.43^)1.86 _ . 

P4 = 0.56 + 0.8 = 47 ' 2 + °' 59?>11 

(47.2 +0.59p n ) 1.36 + 10X0.2 - QO , nK1>l 
Po = P9 = 1 36 + 02 = + Pn 



246 HEAT ENGINEERING 



1.00 



Plb = 2 X 02" = 10 

(43.2 + 0.514p n ) 1.56 . 

P6 = Pl ° = 0.16 + 0.8 + 1.00 = UA + °- 41 Pl1 
(34.4 + o.41 Pl Q(1.00 + 0.8) 

Pii = 2.00 + 0.8 — = 22A + °- 264 ^ n 

22.1 
Pl1 = OT36 = 30 
Pio = P6 = 46.7 
Pb = P9 = 58.7 
Pi = 64.9 
p 8 = 47.4 

P7 = 46.7 X j£g = 125 
2>i 2 = 30 X |g = 14.3 

This method could be used for triple expansion engines or for 
any arrangement of cylinders. The main point to remember 
is to reduce the equations for any unknown to terms of one un- 
known until at last this term occurs on both sides of the equa- 
tion and the equation gives its value. 

EQUIVALENT WORK DONE BY ONE CYLINDER 

On looking at the cards of Figs. Ill, 112 and 113 it will be 
seen that the total length of the combined diagram represents 
the volume of the low-pressure cylinder and if the intermediate 
lines were removed the card would appear as a single cylinder 
card with very early cut-off. The cylinder would be of the same 
size as the low-pressure cylinder. Hence it may be said that if 
the same initial pressure is used in a cylinder of the size of the 
low-pressure cylinder of a multiple expansion engine as is used 
in the high-pressure cylinder, and if this has the same total ratio 
of expansion, then it will develop the same horse-power as the 
multiple expansion engine. 

It is well to note what differences occur. With the single 
cylinder the whole clearance surface of the large cylinder is 
subject to the full temperature range which would cause excessive 
condensation unless it is of the Stumpf form of cylinder. The 
large cylinder would be subject to full high pressure not only 
requiring heavy walls for the cylinder but causing the rods, pins 



MULTIPLE EXPANSION ENGINES 247 

and bearings to be excessive. The cut-off on the single card oc- 
curs at about one-tenth stroke, while on each of the separate 
cards the division of the expansion caused the pressure to be 
continued to nearly the middle of the stroke on a smaller area 
giving a more gradual curve of turning moment. 

On looking at the computations made above it will be seen 
that if the receiver volume had been made very large there would 
have been little change in the intermediate lines and were the 
receiver infinite in volume these lines would be horizontal. Such 
an assumption is usually made in determining the preliminary 
relative sizes of the cylinders of a multiple expansion engine and 
in addition the effect of clearance and compression are omitted 
in this work. Although in practice the receiver is not more 
than 1}^ times the volume of the cylinder which discharges 
into it, it is formed at times by the large exhaust pipe between 
cylinders. The size seems to have no effect on the operation. 

DETERMINATION OF RELATIVE SIZES OF CYLINDERS 

The combined card without clearance for a i triple expansion 
engine with infinite receivers takes the form shown in Fig. 115. 
The whole card which represents the entire work of the engine 
could be developed alone by 
the low-pressure cylinder. The ; 
volume of this cylinder is ab. 
The volume of the intermediate 
cylinder is cd and that of the 
high is ef if the expansions are 
complete to the receiver pres- 
sures. If, however, the vol- 
umes were cd r and ef there 
would be the free expansion gf FlG 115 ._ Intermediate pressures 
in the high and hd r in the inter- with infinite receivers, 

mediate. In many cases free 

expansion is used as it reduces the sizes of the cylinders and 
therefore the sizes of the parts. There is not a great loss of work 
area due to this. 

The receiver pressures are fixed by the points of cut-off of the 
lower cylinders. Thus, with a given cut-off, in the high-pressure 
cylinder, there is a fixed volume of steam and the line ifh of 
the expansion of this steam is fixed. If now in any cylinder 
the volume at cut-off is known this must be on the line of ex- 



Pi 


V 








/N v 






P2 


<^h 






Pi 


d' d" ^ 


^^P 






248 HEAT ENGINEERING 

pansion and consequently the pressure is determined. If the 
cut-off is made later in the intermediate cylinder the point / 
would be further to the right and consequently the pressure 
would have to be lower. This would produce less work on the 
cylinder. Hence the statement is often made that increasing 
the cut-off of a cylinder reduces the work of that cylinder. If 
this is done on the high-pressure cylinder the same result follows, 
as the other cut-offs will be made relatively earlier by this opera- 
tion and the intermediate pressure will rise. 

The whole area of the combined card being the work which 
could be developed by the low-pressure cylinder gives a method 
of determining the size of the cylinder. The size of this cylinder 
fixes the power of the engine. The sizes of the other cylinders 
determine the relative amount of work done by them but do 
not affect the power of the engine. 

The m.e.p. for the whole card is found by assuming pi, pi, and 
R, the total ratio of expansion. 

m.e.p. =/[| (1 + log e R) -p b ]=P (12) 

/ = diagram factor X clearance factor. 
= 0.65 for naval triple. 
= 0.70 for naval compound. 
= 0.80 for Corliss compound and triple land. 
= 0.90 for pumping engine. 

Another form, if p r , the pressure at the end of expansion is 
assumed rather than R, is 

m.e.p. = f[p r (l + log e g) - p»] = P (13) 

From this the size may be found by assuming 2LN and finding 
Fz. 

2PLF t N ,,.. 

lh -P- = -3300T (14) 

Having Fi the areas of the pistons of the other cylinders are 
found from the ratios of 

cd' , ef 
—r and —r 
ab ab 

Thus Fi = F >°i (15) 

F h = F,£ (16) 



MULTIPLE EXPANSION ENGINES 249 

The lengths cd' and ef are found by obtaining the pressures 
p f 2 and p ,f 2 and drawing the figure or computing the lines cd 
and ef and allowing for the distances dd' and //'. 

Thus cd = ab ^ (17) 

Vr 

ef = ab f- 2 (18) 

There are four ways in which to fix the intermediate pressures : 
(a) equal works, (b) equal ratios of expansion, (c) equal tempera- 
ture ranges and (d) assumed ratios. 

(a) For equal work the card is divided into equal areas. 

Area card = D z [p r (l + log e ^) - p 6 ] = F t (19) 

Area 1. p. card = D t [p r (l + log e — ) - p 6 ] = F z (20) 

Area 1. p. card + i.p. card 

= D l [ Vr (l + log, ^) - Vb ] = F 2 (21) 

Area of three lowest cards 

^ D\vr (l + log, 2J) - p»] = F 3 (22) 

For n stages: F t = -F, (23) 

(24) 

(25) 

In each of these there is only one unknown term p 2 . Hence 
these pressures may be found. 

Thus l[p r (l + log, g) - Pb ] = Pr (l+ log e j£)j - p b (26) 
lo * P ' 2 " (n - *) I 1 - f\ + \ l0 ^ g + ^ Vr (27) 

In the same manner: 



^ 


= ^< 


F 2 




F 3 


n 



R 


_ Pi 
Vr 






R 


= n x 


r 2 X r 3 


X 


r 


= VR 






p'> 


= rp r 






P"* 


= r 2 p r ■■ 


= rp' 2 




P2 m 


= r m p r 


= r V r 


-l 



250 HEAT ENGINEERING 

For the mth receiver: 

(6) For equal ratios of expansion the following method is used : 



(31) 
(32) 
(33) 
(34) 

(c) For equal temperature ranges the temperature range 
between pi and pb is found and this divided by the number of 
stages gives the range per stage. If this is added to the lowest 
temperature successively the various temperatures are found 
and from these the corresponding pressures are obtained. 

Tl ~ Th = AT (35) 

m 

T' 2 = T b + AT (36) 

T" 2 = Tb + 2A7 7 = T'% + AT (37) 

T m 2 = T b + mAT (38) 

(d) For assumed ratios of cylinder volumes there is no necessity 
of finding the intermediate pressure. Trie usual ratios of practice 
are: 

Compound engines D h : Di = 1 : 2 to 1 : 7 
Triple expansion D h : Di : Di = 1 : 2.5 : 6.5 

For compound engines a ratio 1 : 4 is quite common. 

When quick response is needed to a suddenly changing load 
the small ratios are used; i.e., the larger high-pressure cylinders 
are used. 

As an example, suppose it is desired to construct a triple 
expansion engine to produce 2000 kw. from a generator at 100 
r.p.m. with boiler steam at 175 lbs. gauge pressure with 80° F. 
superheat, with the pressure at release — 5 lbs. gauge and a 
back pressure of 2 lbs. absolute. 



MULTIPLE EXPANSION ENGINES 251 

m.e.p. = 0.90 [9.7(l + lo ge ^|^) - 2] = 33.0 

• u onnn n/ 1 a 33.0 X 1000 X F t 
i.h.p. = 2000 X 1.6 = 33^ 

Fi = 3200 sq. in. 
d = 64 in. 
2LN = 1000 

_ 1000 K ., Aft . 

L = 2^000 = 5ft - = 60m - 

Intermediate pressures and volumes. 

Method (a) 

logeP'2 = (| - l) [l - |y] + log e (l89.7 X 9.7 2 ) 

= - 0.53 + 3.26 
= 2.73 



H 



p'2 


= 15.4. 










log e p" 2 


-S-JO+M 


189.7 X 9.7 






= - 0.26 + 4.25 












= 3.99 










v\ 


= 52.1 










Fi 


= Fl £ = 320 ° X 


9.7 
15.4 


= 2020; 


di 


= 51 in. 


F h 


= 3200 ~ = 595; 


d h = 


28 in. 






Engine 28-51 and 64 X 60. 










ethod (b) 













« - W = i9 - 5 

r = Vl9^5 = 2.69 
p' 2 = 2.69 X 9.7 = 26.1 
V" 2 = (2.69) 2 X 9.7 = 70.2 
di = 39 in. 
d h = 24 in. 
Engine 24-39 and 64 X 60. 



252 HEAT ENGINEERING 

Method (c) 

7\ = 377.5 + 80 = 457.5 
T 2 = 126.2 

AT = ^ = U0A 

T\ = 236.6; p' 2 = 23.5 

T\ = 347.0; p" 2 = 129.4. 

di = 42 in. 
<4 = 17% m. 
Engine 173442 and 64 X 60. 

Method (d) 

Assume ratio of 1-2-6. 
di = 45 in. 
dh = 26 in. 
Engine 26-45 and 64 X 60. 

In the above solutions the expansion was complete in the upper 
two cylinders. If the diameters are made slightly smaller there 
will be free expansion in each cylinder. 

The relative size could now be found with the clearance which 
varies from 10 per cent, or 15 per cent, in slide valve engines to 
5 per cent, in ordinary Corliss engines and 2 per cent, in engines 
with the valves in the heads. The events could be assumed and 
the volumes of the receivers after which the cards similar to those 
in Fig. 113 could be computed. 

The pressures of 90 to 120 used with simple expansion engines 
are changed in multiple expansion engines to from 150 to 225 
lbs. gauge. The higher pressures are used with triple expansion 
engines. 150 to 175 lbs. would be used for compound engines 
although higher pressures are used with such. At times lower 
pressures are used with these two stage engines but generally 
lower pressures do not give sufficient expansion. The higher 
pressures should give higher efficiencies but the troubles from 
initial condensation and free exhaust exist here to a great 
degree. In most cases, however, neither of these effects is so 
prominent as with single cylinder engines and the steam consump- 
tion curves are more nearly flat, giving considerable range 
without much variation in steam consumption. Such steam 
consumption curves are valuable in any engine or turbine as 
they give a good efficiency over a great range. 

The back pressure should be reduced to as low a value as 



MULTIPLE EXPANSION ENGINES 253 

will give a gain on the entropy diagram. Owing to the free 
expansion in these engines there may be a condition for which 
an increase of vacuum means a loss. About 28 in. is sufficient 
for the vacuum to be carried on a multiple expansion engine. 

In these engines as in the case of the simple engines, the large, 
slow-speed engines are those which give the highest efficiency. 
The larger the cylinder, the more efficient the engine. 

Method (a) is the one usually employed to fix the relative 
size of cylinders. 

JACKETING 

In these engines the effect of jacketing is not always a source 
of economy except in small sizes. The gain on a triple expan- 
sion engine is probably larger than that found on a compound 
engine or on a simple engine due to the fact that heat added 
to the exhaust steam by the jacket when the moisture on the 
cylinder walls is vaporized is of use in the lower stages. 

A number of tests have been made on large pumping engines 
and it has been found that the engine would consume about the 
same amount of total steam with or without jackets. 

REHEATERS 

Steam as it leaves the cylinder of an engine is likely to be quite 
wet and hence the moisture deposited on the walls at entrance 
into the next cylinder would be increased and this might lead to 
excessive condensation. To cut down this moisture, the receiver 
is equipped with a reheating steam coil containing high-pressure 
steam, the receiver is now called a reheating receiver or reheater. 
The value of the reheater lies in the fact that dry steam is carried 
into the cylinder. The value of this apparatus is also questioned 
as some engines on test show an increase of total heat when 
reheaters are used. Other engines show a decrease in the total 
heat. The heaters should contain enough heating surface and 
have a high enough steam temperature to completely dry the 
steam and to superheat it sufficiently to bring the steam to 
a dry condition at cut-off. If this can be done, a gain should 
result. 

In both jackets and reheaters the apparatus should be thor- 
oughly drained and the supply of steam should be taken to the 
apparatus alone. The steam should be hotter than the steam to 
be heated. The use of steam on its way to the first cylinder for 



254 



HEAT ENGINEERING 



a jacket is bad as this may introduce moisture into the cylinder. 
The use of exhaust steam for a jacket is condemned. The supply 
of steam should be through pipes of ample size. This apparatus 
should always be air vented. Heavy lagging of non-conducting 
material has been found of value in engines in cutting down 
radiation. 

GOVERNING 

The multiple expansion engines are governed by changing 
the point of cut-off on the high-pressure cylinder alone or by 
changing it on each cylinder at the same time and in the same 
ratio. 

In Fig. 116, the method of regulation by changing the cut-off 
on the high-pressure cylinder is shown while the method by 




Fig. 116. — Method of governing by 
changing cut-off on high pressure 
only. 



Fig. 117. — Method of governing 
by changing cut-off proportionately 
on each cylinder. 



changing the cut-off on each cylinder is shown in Fig. 117. In 
this latter method the ratios of the V's are all equal if the pressure 
in the receiver is to remain constant. 



V_ 
V 



Vi 
7'i 



V2_ 



In the first method the receiver pressure changes from a to a' 
and from b to b r . Before the receiver pressure can drop the work 
done must be represented by the curves shown in Fig. 118. In 
this it will be seen that the governor has reduced the area by 
the shaded amount and if this is enough to care for the change in 
load, the point of cut-off will have to gradually change. The 
receiver pressures gradually change to their final values and the 



MULTIPLE EXPANSION ENGINES 



255 



expansion curve takes the position shown dotted in Fig. 118; 
the area between the original and final curves being equal to the 
shaded area. Since this change does not occur at once with re- 
ceivers of any size the action is sluggish. Of course this change 
is accomplished in ten or fifteen revolutions but it will not be 
as steady as that resulting from the method shown in Fig. 117. 
With small fluctuations either method is good. The method of 
Fig. 116 cuts down the work on the lower cylinders leaving the 
high pressure about the same while the method of Fig. 117 
affects all in the same way if not by equal amounts. 

Throttling would be shown in Fig. 119. In this the pressure 
in the receiver would change and although the total work is 




Fig. 118. — Sluggish action due to cut- 
off change on high-pressure cylinder only. 
Shaded area shows decrease of work on 
first stroke. 



Fig. 119. — Governing by 
throttling. 



less, the reduction is mainly on the low-pressure cylinder, as 
there is little change on the high. This method throws out 
the equality of works. 



BLEEDING ENGINES OR TURBINES 

The use of low-pressure steam for heating water, for use in 
warming buildings or for other technical applications is carried 
out in plants having high-pressure steam, by throttling the steam 
through a reducing pressure valve to the lower pressure. Now 
although throttling does not mean a loss of heat, there is the 
loss of available energy if by this is meant the ability to be turned 
into mechanical work. This may be seen by remembering that 
it is a difference in pressure which must exist to give ability for 
work in a cylinder. The production of high-pressure steam is 
usually more expensive than that of low pressure due to the 
higher temperature of the steam ; hence after producing this high- 



256 



HEAT ENGINEERING 



pressure steam many engineers believe that it should be brought 
to its low temperature by passing it through an engine or turbine 
where its available energy may be utilized. After reaching the 
desired temperature and pressure this steam may be used. If 
the engine or turbine is connected to a 
condenser, provision is made to take off 
a portion of the steam between stages 
where the pressure is at or near the 
atmospheric pressure. Were this amount 
constant the machines could be made 
to be operated with one quantity to one 
pressure and with another quantity to a 
lower pressure. The card from such an 

application on a compound engine is 
Fig. 1 2 —Combined Jf . p . 1on ^ , & , 

cards from a compound shown m Fig. 120. The volume a has 
engine with steam removed been taken for use outside of the engine, 
from receiver. Cylinders mi ^ r j u u 

of equal sizes. •*■ ne low-pressure cylinder has been as- 

sumed to be similar to the high-pressure 
cylinder in this case. If no steam is taken the low-pressure 
card becomes rectangular as shown by dotted line. 




REGENERATIVE ENGINES 

The preceding suggests the regenerative steam engine cycle 
in which steam is taken from the receiver to raise the feed water 
to that temperature before 
entering the boiler. Steam p ^ 
from one receiver brings the 
water to that temperature and 
that from the next receiver 
brings the water and the 
steam condensed from its tem- 
perature to the temperature of 
the next. To see the applica- 
tion of this suppose that it is 
applied to a triple expansion 
engine the diagram of which 
is shown in Fig. 121. The 

decrease in the volume of each card is due to the removal of 
a certain amount of steam to mix with the feed water to raise 
it to the temperature of the receiver. There would be two 
analyses of this depending on whether or not the condensed 




Fig. 121. — Combined cards from a re- 
generative triple expansion engine. 



MULTIPLE EXPANSION ENGINES 257 

steam from the feed heaters is pumped into the feed. Of course 
this should be done. The expansion is assumed adiabatic in 
theory and 1 lb. of steam will be assumed on the lower card. 
The pressures are p h p 2 , pz, Pa and p . 

Heat to raise 1 lb. of water from t to tz = q z — q 

Amount of steam at quality x z to give this = — = M 2 

#3?*3 

Amount of steam on second card = 1 + M 2 

Amount of steam of quality x 2 to raise 1 + M 2 lbs. of water 

from U to t 2 = — = Mi 

x 2 r 2 

Amount of steam on first card = 1 + M 2 + M± 

Heat to raise 1 + M 2 + M± lbs. of water from t 2 to ti = 

(1+ > M 2 -^M 1 )(q , 1 -q f 2 ) 

Heat to vaporize (1 + M 2 + M x ) lbs. = (1 + M l + M 2 ) Xin 
Total heat from outside = (1 + M 2 + M^Ih - q 2 ] = Qi 
Work done on cycles = (1 + M 2 + ikf i)U'i - i 2 ] + (1 + M 2 ) 

Hi - u] + 1{U' 3 - u] + A(p 4 - Vo)vi\ = AW 

The i's are found on the same adiabatic for 1 lb. of steam from 

one end to the other. 

AW 

Efficiency = -yy— 

This is compared with efficiency for the same weight in each 
cylinder. 

Efficiency = h - h + A(p. - p.)v< (39) 

and the saving is shown. 

Let this be applied to a compound engine working between 
150 lbs. absolute and 2 lbs. absolute with original dry steam 
and a receiver pressure of 45 lbs. absolute and a pressure at re- 
lease in the low of 8 lbs. absolute. 

ii = 1192.6 q\ = 330.0 

i 2 = 1100.0 q 2 = 243.7 

h = 987.0 q' 3 = 150.9 

v 3 = 39.9 q\ = 94.2 

x 2 - 0.92 r 2 = 927.5 
243.7-94.2 
Ma = 0.92X927.5 = °- 176 

17 



258 HEAT ENGINEERING 

Heat = 1.176[1192.6 - 243.7] = 1113 

Work = 1.176[1192.6 - 1100] + l{ll00 - 987 + 

(8 - 2)39.9} 

= 108.7 + 157.4 = 266.1 
266.1 



144 

778 



Theoretical eff. 



1113 



= 0.236 



Theoretical eff. of ordinary cycle = 



144 

[1192.6-987+^(6X39.9)] 



250.0 
1098.4 



1192.6 - 94.2 



= 0.228 




LP.304b.sp. 

60x66 




Fig. 122. — Individual and combined cards from test of pumping engine. 



This means a saving of 0.8 in 22.8 or 3% per cent. It will be 
seen from the expression for the work on each cycle that the low- 
pressure cylinder is doing 50 per cent, more work than the high- 
pressure cylinder. 

Data from the test of a triple expansion engine with its cards 
given in Fig. 122: 



MULTIPLE EXPANSION ENGINES 259 

Size of pump 20,000,000 gal. per 24 hr. 

Size of cylinders. 

Steam 30 in., 60 in. 90 in., X 66 in: 

Water 33 in. X 66 in. 

Pressures by gauge 

Steam supply 180. 2 lbs. per sq. in. 

1st Receiver 25.4 lbs. per sq. in. 

2d Receiver 9.5 in. vacuum 

Condenser 26 . 7 in. vacuum 

Water discharge 86 . 9 lbs. per sq. in. 

Barometer 14.86 lbs. per sq. in. 

Revolution in 24 hr 28,989 

Displacement per revolution. 732 gal. 

Temperatures. 

Return from condenser 113° F. 

Water pumped 40° F. 

Condensing water 40° F. 

Water leaving condenser 70° F. 

Outside air 30° F. 

Quality of steam . 989 

Water pumped to boiler, per hour 9,266 lbs. 

Boiler leakage, per hour 196 lbs. 

Drip in engine room, per hour 135 lbs. 

Steam to drip pump, per hour 58 lbs. 

Net boiler steam, per hour 8,877 lbs. 

Jacket drip, per hour 661 lbs. 

1st Receiver drip, per hour 422 lbs. 

2d Receiver drip, per hour 361 lbs. 

Indicated horse-power steam end 

High pressure 332 . 02 

Intermediate 269. 10 

Low pressure 260 . 22 

Total 861.34 

Water end 839 . 8 

Delivered horse-power 822 . 9 

Steam per i.h.p.-hr 10 . 37 lbs. 

Heat per i.h.p.-min 193 . 04 B.t.u. 

Duty per 1000 lbs. dry steam 181,000,000 

TESTING AND ANALYSIS 

There are a few points which must be brought out in regard to 
testing multiple expansion engines. 

The exhaust from these engines is not all at the temperature of 



260 HEAT ENGINEERING 

the condensed steam when reheaters or jackets are used. In 
such a case the drips from the jacket or reheater are at tempera- 
tures much higher. These possess the temperature correspond- 
ing to the steam pressure. Hence if M pounds of steam are used 
by the main engine, and m,- and m r are the amounts used in 
the jackets and reheaters, the amount of heat chargeable to the 
engine is 

Qi = (M + m,- + m r )ii — Mq — m^qj — m r q ' r (40) 

In this q'j and q' r are the heats of the liquid in the jacket and 
reheater. 

The efficiency which has been given as 

AW 

77 " (M + mj + m r ){i-q' ) m) 

AW 

is nOW 7] = /7l/f - , : r^ Tjr-, -, — (42) 

(M + rrij + m r ) i — Mq — m,jq ■ ,■ — m r q r 

In Hirn's analysis for multiple expansion engines a new device 
must be used to find the heat delivered to the second cylinder 
and discharged from the first cylinder. In analyses of multiple 
expansion engines it is necessary to find Q r , the heat radiated from 
each cylinder. Having this for the first cylinder, the heat losses, 
Q a , Qb, and Qd, are found by the formulae and then Q c is given by 

Qc = Qr~ (Qa + Q b + Qa) (43) 

The heat exhausted from the first cylinder is 

Q 2 = AU* - AU 3 + AW a - Q c (44) 

or Q2 = Qi-Q r - A(W a + W b + W c + W d ) (45) 

The heat given to the second cylinder Q', is 

Q'i = Q 2 + Q h - Qi (46) 

Qh is the heat given to the steam in the reheater and Qi is 
the heat lost in radiation from the reheater. This term Qi is 
found by a radiation test similar to that made on the engine. 

By observing the condensation from the receiver coil when no 
steam is passing from one cylinder to the other, the value of 
Qi is found while the observations during the operation of the 
engine gives the term Q h . 

Q h or Q r = M(i - q f ) (47) 



MULTIPLE EXPANSION ENGINES 261 

Equation (45) furnishes the heat supplied the second stage and 
a similar procedure is used for the lower stages. 

If a jacket is used on any cylinder, the heat supplied by the 
jacket is found by weighing the steam condensed. The heat is 
given by 

Qi = M(i - q') (48) 

If an engine is jacketed there is a new term in Q r : 

Qr = Qa + Qb + Qc + Qd + Q, (49) 

and for multiple expansion engines Q c is given by 

Q c = Qr - (Qa + Q b + Qd + Qi) (50) 

Then Q 2 = Q 1 + Q, - Q r - A(W a + W b + W e + W d ) (51) 

BINARY ENGINES 

Another method of utilizing the waste heat and avoiding the 
waste due to free expansion on account of the great volume of 
low-pressure steam is to use the exhaust to volatilize a liquid, 
such as sulphur dioxide, which has a much higher saturation 
pressure than steam. In this way the same limits of temperature 
may be utilized without passing to such low pressures as used on 
steam engines. Engines using two vapors are known as Binary 
Engines. In a development of this engine by Professor E. Josse 
of Berlin, steam from an engine was exhausted into a surface con- 
denser in which the condensing fluid was volatile SO2, which by its 
evaporation removed heat from the steam and condensed it. The 
pressure of the steam was 3 lbs. per square inch. This gave a tem- 
perature of 141° F. S0 2 at 132° F. is under pressure of 132 lbs. 
per square inch so that the S0 2 would boil and produce a pressure 
of 132 lbs. absolute while condensing steam at 3 lbs. pressure. 
This could be used in a cylinder connected with the same shaft as 
the steam cylinders and aid in the driving. This cylinder ex- 
hausted into a condenser cooled by water at 50° F. in which 
the S0 2 condensed at a pressure of 31 lbs. absolute and a tempera- 
ture of 66° F. The condensed S0 2 was pumped into the steam con- 
denser where it was evaporated again and used. 

In the steam engine Josse developed about 150 h.p. on 250 
B.t.u. per horse-power minute and by using the exhaust as de- 
scribed above, 50 additional horse-power were developed giving 
176 B.t.u. per horse-power minute. 



262 HEAT ENGINEERING 

The reason for the gain in the binary engine is found in the 
fact that the effect of initial condensation on a large cylinder 
or the effect of free expansion have been eliminated. These 
are practical reasons and not theoretical. The range of tempera- 
ture has not been increased but the pressure and volume limits 
have been so changed that these engines have a high practical effi- 
ciency, TJi. 

Josse has installed the engines commercially and has found that 
they are valuable provided a high load factor can be had. For 
an intermittent load or for use at intervals the engine is too 
expensive. 

TOPICS 

Topic 1. — What are multiple expansion engines? Why were they used? 
Why are they of value? Explain how the combined diagram is constructed. 
What is the difference between the Woolf and the receiver compound 
engines? 

Topic 2. — Sketch the combined cards for a Woolf compound engine. Write 
the formulae by which the pressures at the various points may be computed. 
Explain why 

VaVa +PvVv 

;■ = Vr. 

Va + V v ^ 

Topic 3. — Sketch the combined cards for a receiver compound engine. 
Write the formulae by which the pressures at the various points may be com- 
puted. Explain why 

PaVa+ PvVv _ p 
V a + V v 

Topic 4. — Explain why it may be said that the size of the low-pressure 
cylinder fixes the power of a multiple expansion engine. On what does the 
size of the higher pressure cylinders depend ? If the size of the high-pressure 
cylinder is changed what is the effect of this? 

Topic 5. — Derive the formulae for the total work of an n-stage engine and 
for the work on the low-pressure cylinder. From these find the pressure at 
entrance to the low-pressure cylinder. 

Topic 6. — Give the method of finding the intermediate pressures for equal 
ratios of expansion, equal temperature ranges and given ratios of volume. 
Knowing the pressures, show how the volumes of the various cylinders are 
found assuming (a) complete expansion and (b) some free expansion. What 
is the reason for assuming receivers of infinite capacity? 

Topic 7. — Why is a 28-in. vacuum a limit for vacuum in steam-engine 
operation? What is the effect of jackets? Reheating? Explain what 
happens when the cut-off is made later on any cylinder and the reason for 
this effect. 

Topic 8.— Discuss with diagrams the various methods of governing multi- 
ple expansion engines and show which is the better form. 



MULTIPLE EXPANSION ENGINES 263 

Topic 9. — Explain what is meant by bleeding engines or turbines and why 
this is of value. Explain the action of the regenerative engine. Give the 
expressions for the various amounts of heat and work and derive the expres- 
sion for efficiency. 

Topic 10. — Derive the expression for the heat delivered to one of the lower 
stages of a multiple expansion engine (Q\ or $2of Hirn's analysis). Show 
what effects jackets or reheaters have on this formula. What heat is charge- 
able to an engine when the drips from the jacket and receivers may be sent 
back to the feed ? What is a binary engine ? 

PROBLEMS 

Problem 1. — Find the pressures at the corners and events of the stroke 
for a compound engine, 8 in. and 18 in. X 24 in. — 80 r.p.m., with a re- 
ceiver of volume equal to twice the volume of the high -pressure cylinder 
and with 8 per cent, clearance on the high-pressure cylinder and 4 per cent, 
on the low. Initial gauge pressure is 125 lbs. and the back pressure is 2 
lbs. absolute. Cut-off is 25 per cent, and compression is at 20 per cent, of 
the stroke from the end. Cranks at right angles. 

Problem 2. — Construct the combined cards for Problem 1 to scales of 1 in. 
= 30 lbs. and 1 in. = 0.435 cu. ft. Find the m.e.p. of each card. Construct 
the h.p. card to scales of 1 in. = 60 lbs. and 1 in. = 0.1745 cu. ft. and the 
l.p. card to scales of 1 in. = 20 lbs. and 1 in. = 0.884 cu. ft. Find the 
scale of B.t.u. per square inch of area of card, and of ft.-lbs. per square inch 
of card. Find the h.p. of the engine. 

Problem 3. — Find the equivalent m.e.p. to be expected in Problem 1 if 
the complete ratio of expansion is 10. What is the size of the low-pressure 
cylinder to develop 350 i.h.p.? 

Problem 4. — Find the size of the low-pressure cylinder of a compound 
Corliss engine to drive a 2000-kw. generator. The initial pressure is 160 
lbs. gauge, the back pressure is 3 lbs. absolute, and the pressure at end of 
expansion in the low-pressure cylinder is —4 lbs. gauge. Find the size of the 
high-pressure cylinder by four methods. 

Problem 5. — Find the receiver pressures for a triple expansion engine 
operating between 175 lbs. gauge and 2 lbs. absolute, with a complete ratio 
of expansion of 22. 

Problem 6. — Construct a p-v diagram for Problem 5 with infinite receiver 
and no clearance and compression. Construct the T-S diagram for the same 
cards and show whether or not it would pay to reduce the back pressure to 
1 lb. absolute. 

Problem 7. — Find the sizes of cylinders to be used with a compound engine 
developing 1000 h.p. if the steam consumption of the engine is 15 lbs. per 
1 h.p.-hr. when 30 per cent, of the steam is removed from the receiver and 
the engines develop equal works. The limits of pressure are 150 lbs. gauge 
and 3 lbs. absolute. 

Problem 8. — Find the thermal efficiency of a triple expansion regenerative 
engine with gauge pressures as follows: Initial high, 170 lbs.; first receiver, 
65 lbs.; second receiver, 5 lbs.; end of expansion, — 8 lbs.; back pressure, 
- 12 lbs. 



264 HEAT ENGINEERING 

Problem 9. — Design a cylinder to deliver 120 i.h.p. at 100 r.p.m. with 
pi = 130 lbs. gauge, p v = 3 lbs. gauge, r = 5. Find the probable x at cut- 
off. Find the probable steam consumption. Find the pressure of SO2 
evaporated by the heat of the exhaust. Find the back pressure in an engine 
working with this S0 2 if the condensing water is 70° F. Sketch a card for 
the SO 2 cylinder. Find the volume of SO 2 vapor per minute at release and 
from this find the horse-power of the S0 2 cylinder and its size at 100 r.p.m. 
What is the thermal efficiency of the combined engine? At for heat trans- 
mission is 15° F. 



CHAPTER VII 
STEAM NOZZLES, INJECTORS, STEAM TURBINES 

NOZZLES 

In Chapter I the formula for the velocity of discharge of any 
fluid through a passage from pressure pi to pressure p 2 was de- 
rived. This gave 

w 2 = \/2gJ(i! - i 2 ) = 223.7\A'i - k C 1 ) 

ii and i 2 are the heat contents at the two pressures regardless 
of whether or not there is internal friction but dependent only on 
the fact that there is no external heat transfer. It was pointed 
out that i 2 for any pressure was difficult to find if there was in- 
ternal friction and hence the ordinary method was to find the 
value of i 2 on a reversible adiabatic (which means no friction) 
through pi iij and to subtract from the heat (i\ — i 2 ), which 
should have been transformed into kinetic energy, the amount 
utilized in friction. This gives the amount remaining for change 
in kinetic energy. If y{i x — i 2 ) is the amount of energy used in 
internal friction against the sides of the passage and between the 
particles of the fluid, the amount left for the change of kinetic 
energy is 

(ti-«(l -y) (2) 

This quantity represents the change in kinetic energy and gives 
w 2 = 223.7 Vfe - *" 2 )(1 - y) (3) 

when if i is so small that ^— may be neglected or 



v>2 = 223.7^-^(1 -V)+^f ( 4 ) 

when Wi is appreciable. If w\ is 200 ft. per second it is inap- 

A /iii 2 

preciable as — ^ — is 0.8 while the first term of the bracket may 

be from 50 to 100 or even 250 B.t.u. The importance of this 
term is determined by the conditions of any problem. 

265 



266 



HEAT ENGINEERING 



The value of y is determined by experiment. w 2 is measured 
by the reaction of the jet and weight of the steam or by a pitot 
tube and from this measured value of w 2 , the value of y is com- 
puted. By measuring the quality of the steam actually present 
at 2 and comparing it with that of adiabatic expansion the value 
of y may be found. In most cases y will vary from 0.10 to 0.15 
for long nozzles with large angles of divergence or for length of 
10 diameters for small angles of divergence to 0.06 or 0.08 for 
shorter tubes. It seems that with nozzles the coefficient de- 
creases with the increase of pressure. With orifices in plates the 
value of this y may be 15 per cent. 

Suppose that the pressure drop is uniform along a tube and that 
frictional loss is zero. If the velocity at any point be computed 

by 

w x = 223.7\/(ii - ix) (5) 

and 1 lb. of steam is assumed to be passing per second 
the area may be computed by 

F = M l^ = Mxv "* = "iM. 

w x w x 4 



(6) 



The values at different points for the above assumptions are 
tabulated below. 

Nozzle Data 



Pressure 
absolute 


Quality 


Heat 
content 


Specific 
volume 


Velocity 


Area, 
sq. in. 


Diam., 
meters 


149.1 


0.4° sup. 


1193.3 


3.04 








124.4 


0.987 


1178.3 


3:55 


865 


0.592 


0.87 


100.2 


0.972 


1161.0 


4.30 


1270 


0.487 


0.79 


75.4 


0.953 


1138.8 


5.51 


1640 


0.484 


0.78 


50.0 


0.929 


1108.1 


7.91 


2060 


0.554 


0.84 


24.97 


0.894 


1059.3 


14.58 


2590 


0.810 


1.02 


14.99 


0.872 


1026.0 


22.91 


2890 


1.140 


1.20 


4.97 


0.830 


959.9 


61.20 


3410 


2.580 


1.81 



The diagram of Fig. 123 will be obtained, in which velocity, 
specific volume, area, and radius are plotted. It is to be re- 
membered that in this figure the pressure drop has been made 
uniform. The values of i\ and i 2 were found in the same entropy 
column and x x v" x was found in that column or computed from 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 267 



the value of x. It will be seen that the required area reaches a 
minimum section and then increases. This is due to the fact that 
the specific volume curve does not increase as rapidly as the ve- 
locity until the critical point of discharge is reached and after 
this the more rapid increase of specific volume makes it necessary 

150 

o 

I* 75 

o 

3000 



2000 



1000 



15.0 



10.0 



u 

£ 5.0 



1.5 

* 1.0 

"■ 0.5 



S 0.75 

*0.5 

0.25 













^J»'e Pnil . 




















r^as 


^£Sq 


to. 


































t.-f et * 


,eco^ 
















VeV> 


Uyj^ 










Lh 






















/ 




1 


20 


c 





C 





3 


y 
















"? 


^'S 












Spec 


.&c^j 


A^ 


(£>■ 






I 






















1 




















\ 


















/ 


\ 




^ A 


~nea_ 


-Fass 


age * 


d.^ 














































I 


















^ 


J 








Radii 


of Nc 


zzlel*^ 8 





































Length 



Fig. 123. — Curves of velocity, volume, area and diameter for nozzle 
assuming uniform pressure drop. 

to increase the area. This critical point is at the pressure point 

(2 \ n 
— j—rjn-i pi = 0.57pi for steam (7) 

when n = 1.135. 

Whenever p 2 is less than this critical value there must be' a 
minimum section in the tube. Such a tube is spoken of as a 



268 



HEAT ENGINEERING 



nozzle and the small section is known as the throat and the large 
end is known as the mouth. On account of the high pressure 
available at the entrance to a nozzle the section from the entrance 
to the throat is made very short. The length of this is usually 
made about equal to one or two times the diameter of the throat. 
A nozzle may be assumed of the section shown in Fig. 124, and 
the areas at various points are found and plotted. At various 
assumed pressures the velocity is determined and then the area 
to carry 1 lb. of steam is computed by the formulae: 



w 2 = 223.7\A*i - k 
F 2 = 



x 2 v" 2 



w 2 



(1) 

(6) 



The position on the curve of area for this value will fix the posi- 
tion of the assumed pressure. Following this for the various 
points the curve of pressure is obtained. The following table 
gives the computation for the curve, using Peabody's Entropy 
Table for entropy 1.56 and running from 160.8 to 14.7 lbs. 
pressure. 



Pres- 


160.8 


121.1 


90.9 


50.0 


30.35 


20.02 


14.7 


10.17 


sure 


















ii 


1191.2 


1191.2 


1191.2 


1191.2 


1191.2 


1191.2 


1191.2 


1191.2 


ii 


1191.2 


1168.1 


1145.4 


1100.7 


1065.5 


1037.7 


1018.0 


995.2 


i\ — U 





23.1 


45.8 


90.5 


125.7 


153.5 


173.2 


196.0 


w 2 





1071.0 


1510.0 


2120.0 


2500.0 


2780.0 


2940.0 


3130.0 


X 2 V" 2 


2.811 


3.601 


4.63 


7.84 


12.18 


17.60 


23.13 


32.08 


Area 


00 


3.36 


3.06 


3.70 


4.87 


6.12 


7.87 


10.2 


-f- ia 3 



















At 0.57 of 160.8 or 91.7 lbs. the area is 0.00306 sq. ft. 
Napier's formula this area would have been 

70 X 1 



By 



160.8 X 144 



= 0.00302 sq. ft. 



a result practically the same as that in the table above. In 
drawing the nozzle of Fig. 124 the area at the end has been made 
0.00985 sq. ft. which is larger than 0.00787 sq. ft. required at 14.7 
lbs. pressure. This is known as overexpansion and the steam 
will expand to below the pressure at the end and be brought up 
again as shown in the figure. It means a loss in efficiency. This 
effect is greater than the loss due to underexpansion as will be 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 269 

seen later. The reverse curve near the throat indicates that the 
distance from entrance to throat or minimum section is too long. 
A shorter length giving the steeper curve would be better. 

If now friction be assumed and the areas be found for different 
pressures by the formula} : 

w 2 = 223.7\/ft 



ii){i -y) 



F 2 = 



w 2 



(4) 
(6) 



the curve of pressure drop with friction may be had for any given 
form of nozzle in the same manner as that used above. In 
formula (6) the value of x' 2 v" 2 is not found in the same entropy 



0.0010 



0.005 



150 



100 



50 



















T 


abe or No 


zzle Secti 




on 














V 


*5 








V- 


^" 
































£^2^e 





Fig. 124. — Pressure curve of nozzle obtained from velocities and areas. 

column or line as that on which i\ was found because in this case 
the heat content at discharge is i 2 of the isoentropic line for the 
pressure p 2 plus the friction y(ii — i 2 ). If curves for various 
values of y are computed and these curves are compared with the 
actual pressure at various points, the probable values of y may 
be found. 

Stodola has explored the different points of a nozzle by an 
exploring tube introduced along the axis. This tube was closed 
at the inner end and connected to a pressure gauge at the outer 



270 



HEAT ENGINEERING 



end. Holes were bored into this around the circumference at 
one point in the length. In place of being normal to the tube or 
at right angles to the axis they were inclined at 45°. There were 
two tubes, one with the holes inclined in the direction of flow 




Fig. 125. — Stodola's apparatus for exploring nozzles. 

and one with them inclined against the direction of flow. The 
readings of these varied, due to the impact of the steam increasing 
the pressure in one and decreasing it in the other. The mean 
value was used. In addition normal holes were introduced into 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 271 

the side wall of one tube as shown. The pressures at the large 
end of the nozzle showed that for this part the curve of the value 
of y = 0.10 agreed closely with the actual curve although there 
was a slight increase in y as the end was reached. 

Stodola found in this experiment that the pressures at the wall 
of the nozzle were practically the same as those at the center of the 
nozzle showing that with the conical nozzle there was pressure 
exerted on the wall confining the steam. If there were no wall 
present as in the case of discharge from a hole in a plate the 
steam would be forced outward and would follow a curved path, 




■ 100 lbs. 




Fig. 126. — Stodola's exploration 
curves for nozzles. 



Fig. 127. — Stodola's exploration 
curves for orifices. 



the centrifugal force exerting the necessary radial pressure to 
allow the steam to have an acceleration axially. This cen- 
trifugal force is only another way of expressing the inertia of the 
steam against radial acceleration. 

If a valve is put in the pipe line beyond the nozzle, the pres- 
sure of the discharge region may be raised, and the curve found 
by the exploring tube shows what may happen in over expan- 
sion. Curves obtained by Stodola are shown in Fig. 126. 

The waves set up at low pressures are due probably to acous- 
tic vibrations which are so dampened by friction as to be elimi- 
nated in a short time. This is due to the fact that the velocity at 



272 



HEAT ENGINEERING 



the critical pressure of the throat corresponds to the velocity of 
sound. This velocity of course depends on the pressure and tem- 
perature of the substance. Stodola points out that this is an 
illustration of Riemann's Theory of Steam Shock. 

If an investigation by the exploring tube is made on a hole 
with a rounded entrance, known as an orifice, shown in Fig. 127, 
curves of somewhat similar forms are found and it will be noted 
that the pressure in the plane of the orifice is practically constant 
although it reaches a lower value beyond the face of the orifice. 
With a sharp-edged orifice the same observations were made 
although here the pressure at the outer plane of the orifice seemed 
to be the critical pressure. The crossing of the pressure lines at 
the mouth of the orifice means that the velocity at this point is 
the same whatever be the outer pressure, hence the quantity 
discharged will be the same whatever the lower pressure is, pro- 



IP 




m 




gjjl^ 




mm 






Fig. 128. — Mouthpiece or orifice. 



ms U m 



Fig. 129. — Rateau's orifices. 



vided it is less than the critical pressure. Above this the line of 
pressure at the crossing of the plane of the orifice is greater and 
varies with the pressure p 2 ; the velocity in the tube is less than 
before. The great drop in pressure in this figure shows that the 
axial distance to the point of critical pressure may be very short 
and hence the arrangement of the nozzle in Fig. 124 was pos- 
sible. It could have been made shorter. The existence of this 
throat demands a greater total pressure drop to the end than the 
amount to bring the pressure to the critical point. Hence a 
nozzle of the form of Fig. 124 would mean excessive overexpan- 
sion if the pressure p 2 were not below the critical pressure, as 
shown in Fig. 126 with a high back pressure. This would mean 
shock and resultant loss. In the case of a drop in pressure to a 
point above the critical pressure, the expansion part of the 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 273 

nozzle is omitted giving the orifice of Fig. 128. On account of the 
pressure not reaching the critical point in certain cases it may be 
well to stop the curves at points earlier than the section with the 
parallel elements. This would give the forms suggested by 
Rateau, shown in Fig. 129. These are known as orifices or 
converging nozzles. 

In these, the converging portion reduces the steam pressure to 
a point above the critical pressure. Rateau showed that the 
coefficient of discharge of these nozzles varied from 0.94 for a 
small pressure drop to unity at the critical point where p 2 = 
0.57 pi. For the thin orifice shown in the figure the coefficient 
of discharge was 0.82 at its critical point. This means that in 
designing nozzles the first step is to find the relation between 
the pressure p 2 and p\. If p 2 is greater than 0.57pi, the nozzle 
is of the form shown in Fig. 129. That is, it is an orifice or 
mouthpiece. If p 2 is less than 0.57pi there will be a minimum 
section or throat and there is a diverging part, best spoken of as 
a nozzle. 

The velocity and area are computed for the throat of the orifice 
or the throat and mouth of the nozzle. The steps will be best 
illustrated by two problems. 

Suppose a nozzle is to be designed to discharge 10 lbs. of steam 
per second from 160.8 lbs. absolute pressure and quality 
0.9966 into a vacuum of 20". This corresponds to 4.91 lbs. 
absolute. The areas are required. 

p 2 < 0.57pi or 91.66 lbs. 

Hence there is a throat as well as a mouth. 

In figuring the velocity at the throat, the length of this is so 
short for the drop in pressure that the friction is negligible while 
for the main length of the diverging portion there must be a value 
of y as mentioned on p. 266. The y for the problem considered 
with a long tube and large drop is 0.15. For a drop of a little 
less than this from the critical pressure the value of y would be 
about 0.075 if there were a short diverging portion to the nozzle. 
w t = 223.7\/ (1191.2 - 1146.0) = 1501 ft. per sec. 
w m = 223.7V (H91.2 - 953.1)0.85 = 3175 ft. per sec. 

v t = 4.60 
v m is found on the line for 4.91 lbs. pressure but at a value of 
i m equal to 953.1 + 0.15 (238.1) or 988.8. This is found to be 
64.0 at 8 = 1.617. 

18 



274 



HEAT ENGINEERING 



_, 10X4.60 AnQ1 

F t = —[zqi — = °- 031 s q- ft - 

^ = ^§^° = 0.202 sq.ft. 
Moyer states that F m may be found from the formula 
F 2 =Fi [0.172^ + 0.70] (8) 

Applying this 

F 2 = 0.031 [o.172^~ + 0.70] = 0.195 sq. ft. 



This rule would therefore make the nozzle with slight under- 
expansion and the effect of this would be to reduce the velocity 
by the percentage shown by curve of Fig. 130 given by Moyer in 
his Steam Turbines. This figure shows that the effect of under- 



|_5 

>-« 
-0-3 

o 

-5-1 

PL, 



25 20 15 10 5 

% of Under Expansion 



10 15 20 

oi Over Expansion 



25 30 



Fig. 130. — M oyer's curves showing loss due to variation in area of mouth. 

expansion is less than that of overexpansion. In the problem 
above the amount of under-expansion by this simple rule is 

202 - 195 



202 



or 33^ per cent. 



The effect of this is to produce practically no change in the com- 
puted velocity. 

The length of the nozzles is usually four or five times the 
diameter of the throat regardless of the amount of flare. If 
however the flare is over 6 deg. it would be well to make the 
divergent part a curve so as to support the steam pressure and 
accelerate the velocity. With a great flare there is no support 
for the steam pressure and a lower velocity results. 

If the pressure p 2 is 121.1 lbs. absolute, the above computations 
show that this is above the critical pressure and the mouthpiece 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 275 

is to be of the orifice or converging nozzle form. In this case 
the smallest section is at the end. If the length of this orifice is 
made 3d 2 it might be well to assume a friction factor of 4 or 
5 per cent, giving: 



w 2 = 223.7\/(1191.1 - 1168.1)0.96 = 1050 ft. per sec. 

v for 121.1 lb. and i = (1168.1 + 0.92) = 1169 is 3.605 

__ 10 X 3.605 nn _._ 

F 2 = — jggg = 0.0343 sq. ft. 

In figuring the discharge from a nozzle or orifice when p 2 is less 
than 0.57 pi the method is to compute the discharge through the 
smallest area as if the critical velocity existed at that point, the 
velocity being computed for the drop to the critical pressure and 
the volume taken for that point. This method is used regardless 
of how much less p 2 is than 0.57pi. All that is meant is that the 
pressure at this small section is the critical pressure. This is 
always true for orifices while for nozzles a value of p 2 different 
from the one for which F 2 was designed will cause a change in 
velocity due to the effect of under- or over-expansion, although the 
weight discharged will remain constant. 

INJECTORS 

The first application of the discharge from nozzles is to the 
injector. This machine was invented by Henri Jacques Giffard 
in 1858 although the same principle had been applied by others 
for the raising of water by jets. Giffard's injector was intro- 
duced into various countries and through changes suggested by 
its application it was improved by a number of inventors. One 
of the modern forms of injectors is shown in Fig. 131. This is 
the Seller's Improved Self-acting Injector. Steam enters at A 
and to start the apparatus the handle B is drawn back a slight 
distance so that steam may enter the annular space C, the plug D 
preventing any entrance at the center. The combining tube E 
has openings F and G leading into the space H which is con- 
nected to the atmosphere by the valve I which may be held down 
by the cam J. Steam rushing through the small space at the 
end of C produces a high velocity and, due to the fact that the 
space H is at atmospheric pressure since the valve I opens to the 
atmosphere, the pressure at the end of C may be below that of 
the atmosphere due to the over-expansion. This together with 



276 HEAT ENGINEERING 

the entrainment of air by the steam jet gradually sucks the air 
from K producing a vacuum in this space, thus drawing water 
into the chamber. This water meets the steam issuing from C, 
and entering the tube E the water passes through the openings 
into H and finally overflows into the atmosphere as the cam J 
is raised. When water appears at L the handle B is drawn back, 
and steam entering the center imparts such a velocity to the 
water that when the velocity head is changed into pressure head 
by the diverging delivery tube M there is sufficient pressure to 
move the check valve N and allow the water to enter the boiler 
feed pipe 0. 

The function of the steam nozzle C is to lift the water to the in- 
jector while that of the steam nozzle P is to give sufficient steam 




Fig. 131. — Seller's improved self-acting injector. 

to mix with this water in the combining tube E so that a high 
velocity will occur at the throat Q where the steam is completely 
mixed with the water. The function of the diverging delivery 
tube is to reduce this high velocity and change this kinetic energy 
into pressure or potential energy. The valve R opening inward 
is used to allow extra water to enter H and so mix with the steam 
through F and G when the pressure in the combining tube is 
less than atmospheric pressure, due to a lack of water at K, as 
the plug valve S may not be opened sufficiently. An excess 
pressure in H caused by too much water for the steam is shown 
by a slight leakage at the overflow L. Thus the tube C will 
restart the injector and the valves R and I care for a deficiency 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 277 



or excess of water which if not taken account of might cause a 
break in the proper action of the injector. For these reasons this 
is known as a restarting self-acting injector. 

The important parts of the injector are the steam nozzle, 
the combining tube and the delivery tube. The forms of these 
have been the result of much experimentation. The steam noz- 
zle will be discussed first. 

Certain experiments are described by Kneass in his excellent 

treatise, " Practice and The- 
ory of the Injector." One of 
the first things he describes 
on the steam nozzle is a set 
of photographs of their differ- 




135- 
120- 


~\ 


2 

H 








\ ° 




105- 
90- 


1 \ 


a 






Critical 


Pressure 




75- 
60- 




Mouth 




45- 
30- 




V 


o 

5 






Atmosphere 










V'-—^- 







Steam Nozzle 



CombiningTube 



Fig. 132. — Discharge from nozzles 
showing action of steam, according to 
Kneass. 



Fig. 



133. — Exploration curve of 
Kneass. 



ent forms. In the first one, the discharge from a converging 
orifice shows a sudden enlargement at the face and considerable 
disturbance, although this is not so great as that in the dis- 
charge from a diaphragm in a mouthpiece. In this there are a 
great number of cross currents. 

The disturbance in the straight taper is not so extensive and in 
the discharge from the curved divergent nozzle it is practically 



278 HEAT ENGINEERING 

eliminated. The enlargement in the first two jets is due to pres- 
sure which still exists in the steam while in the last two cases this 
pressure has been utilized in driving the steam forward so that by 
the time the end of the nozzle is reached the pressure has been 
utilized. These indicate the form best suited for the purpose. 
To study what takes place in the injector he then explored the 
center of a nozzle and combining tube in the manner described 
before and obtained the pressure curve shown in Fig. 133. In 
this it is seen that the converging part is rather longer than has 
been described but the critical pressure is seen to occur at the 
throat. It will be observed that the end of the orifice is reached 
before the pressure is reduced to atmospheric pressure but this con- 
tinues to fall to a lower point and in the combining tube there is a 
strong vacuum, which is gradually decreased, reaching atmospheric 
pressure before the throat of the delivery tube is reached. With 
no water present the steam pressure, marked by the dotted line, 
shows a drop below atmospheric pressure and then a rise. These 
two curves show that the condensing effect of the water is felt 
but this does not seem to change the pressure curve within the 
steam nozzle as the curves agree almost exactly. 

STEAM NOZZLE VELOCITY 

The form of steam nozzle usually employed is one with a 
converging part followed by a conical diverging part. The areas 
of the throat should be computed to care for the necessary steam 
and the mouth should be figured for atmospheric pressure but in 
order to find the amount of steam for a given amount of water 
the velocity of the steam jet in the combining tube is needed and 
for this the pressure is assumed to be 4 lbs. absolute or a vacuum 
of 22 in. The formulae for the velocities at these points would be 

w t = 223. 7 V i^Jt (9) 

w m = 223.7V \i x - i atm )(l - 0.10) (10) 

w c = 223.7V \ii - H>(1 ~ 0.10) (11) 

iatm — heat content at atmospheric pressure on 

same line as i\ 
it = heat content at 4 lbs. absolute pressure on 
same line as i\ 

WATER VELOCITIES 

The water enters the combining tube under the absolute 
pressure in the suction tank minus the lift, friction head, and 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 279 

the pressure in the combining tube. The friction head is equal 
to a complex quantity due to the sudden bends in the path, the 
sudden changes in section and the friction on the sides. If it is 
assumed equal to twice the velocity head, the following equation 
is true: 

w 2 

=-(1 + 2) = 2.30(pi - Vc) + 34.0 - M 

From this 

w w = V2gy 3 [2.S( Pl - p c ) + 34.0 - hi] (12) 

pi = pressure in suction tank above atmosphere in 

pounds per square inch 
p c = absolute pressure in combining tube taken as 

4 lbs. per square inch 
hi = lift in feet 

34 = feet head equivalent to atmosphere. 
Now the mixture of water and condensed steam in the com- 
bining tube must have sufficient velocity to produce a pressure 
equal to the boiler pressure when reduced to a low velocity of 
about 5 ft. per second. If w m is the velocity of the mixture in 
the combining tube and w b , that on entering the boiler, the formula 
for this velocity would be derived from the Bernoulli equation. 

w m 2 . p c 144 w b 2 . P&144 

-t> h = t> h h losses 

2g m c 2g m b 



Neglecting the losses 



to, 



M(£-_S iu h^ (l3) 

p b == absolute boiler pressure in pounds per square inch 
p c = absolute pressure in combining tube in pounds 

per square inch 
nib = weight of 1 cu. ft. mixture at boiler entrance 
m c = weight of 1 cu. ft. mixture in combining tube 
w b — velocity into boilers, for instance 4 ft. per second. 
By experiment Kneass found that the weight of 1 cu. ft. of 
mixture in the combining tube is about 80 per cent, of the weight 
of water at the temperature of the mixture. This decrease in 
density is due to the presence of air and steam in the mixture. 

m c = 0.80 X m t (14) 

m b = weight of water per cubic foot at t° F. 

= 60 lb. for first approximation 
nib is taken as equal to m t . 



280 HEAT ENGINEERING 

THEORY OF THE INJECTOR 

Now from the above equations (11), (12) and (13) the equation 
for the operation of the injector is obtained. The underlying 
principle of the injector is that of the impact of bodies : The sum 
of the various momenta before impact is equal to the momentum 
after impact. In the impact of the steam and water the action 
is in many directions among the various particles of condensed 
steam and water so that the momentum after mixture is not 
equal to the sum of the individual momenta. Experiment shows 
that it is about six-tenths of the sum or less. An average value 
of 0.5 will be used. If z lbs. of water are assumed to be lifted by 
1 lb. of steam the following momentum equation holds: 

0.5|> c + zw w ] = (1 + z)w m (15) 

This is the fundamental equation of action of the injector. 
Steam on account of its small density and on account of heat 
transfer attains a high velocity from the nozzle and after this it 
is condensed into a more dense substance, water, moving at the 
same velocity. This water may now be allowed to impinge on 
other water moving at a much slower velocity and even after the 
high-speed jet has mixed with eight or ten times its weight of 
slow moving water, the velocity resulting from the impact is 
sufficient to produce a pressure great enough to force the mix- 
ture into the boiler. This will be easily understood when it is 
realized that the steam would issue from a boiler with a velocity 
of 3200 ft. per second while water on account of its greater 
density would issue from the same boiler pressure with a velocity 
of 140 ft. per second. 

HEAT EQUATION OF THE INJECTOR 

There is now enough data for the computation of z since w c , 
w w and w m can all be computed for given conditions. There is 
one quantity which has been assumed and that is m t which was 
made equal to 60 lbs. This had to be assumed for the first approxi- 
mation. After z is found the temperature of the mixture can be 
found by the equating of the energy before mixture to that after 
mixture. 

ix + z(q' w + ^f) = (1 + x) (q' m + ^~f) + losses (16) 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 281 

The energy to be accounted for in each pound of steam is i\ 
before it has acquired any velocity. It must be the same after 
it has attained its velocity since no heat has been added from the 
outside. The energy in the water is that due to temperature 
plus the kinetic energy and the same is true for the discharge. 
These energies are figured above water at 32° F. as a datum plane. 
The terms representing the kinetic energy are so small that they 
may be neglected, giving the energy equation 

i\ + zq' w = (1 + z)q' m (16a) 

From (16) q' m is found and the temperature corresponding 
enables one to find the weight of 1 cu. ft. of water. If this 
differs much from 60 lbs., the value of w m must be recomputed and 
then, a new value of z. Of course the value of 0.8 for the density 
of the mixture is a variable and is not absolutely known in design 
so that a slight variation from 60 need not require recalculation. 

STEAM WEIGHT 

Having z the quantity of steam for a given weight M per hour of 
boiler feed can be found. Here M represents the weight of water 
taken from the suction per hour. 

M 



3600s 



= weight of steam per sec. 



STEAM NOZZLE 



Knowing the weight of steam per second, the area at the throat 
and mouth of the steam nozzle can be found from the formula. 

F = ^ (5) 

Having the areas of the nozzle at the various points the length 
is found by making it taper to the proper area by a taper of 1 in 6. 

COMBINING TUBE 

The combining tube is to be designed in the next step. The 
form of this tube is mainly the result of experiment. The water 
must be sustained as it is struck by the steam and condensed 
steam. The form is a gradually converging tube which is made 
of a length of about 18 or 20 times the diameter at the throat 
of the delivery tube. The end should be slightly rounded and 



282 HEAT ENGINEERING 

the annular area should be equal to that required to admit the 
water with a velocity due to the pressure at this point which would 
probably be about one-half the velocity of the water at the 
point of lowest pressure. 

DELIVERY TUBE 

The delivery tube is next designed. The throat is designed to 
care for the mixture of water and steam and then the tube is 
made divergent so as to cut down the velocity and increase the 
pressure. The velocity at the throat is computed by a formula 
similar to that for w m but different in that the pressure is as- 
sumed to be atmospheric at this point, due to the opening into 
the overflow. 



w' d 



-M*^)™+* (l7) 

The area in square feet required at this point is given by 



3600 



(18) 



m m Aw' d 
A = relative density = 0.8 

This area F d of the combining tube fixes the number of the 
injector as most manufacturers use the diameter of this point in 
millimeters as the nominal size or number of the injector. 

Ft 

Kneass points out that the ratio 77- has a value between two and 

r d 

three. 

The shape of the delivery tube is made in various forms. It 

has been suggested to reverse the 

, — ■ A „„ form advised by Nagle for hose 

1 " ^J nozzles. Nagle suggested that 

^_._ _.. Length _. _._^ ^ e acceleration be made uniform. 

Fig. 134.— Shape for delivery tube. I f the velocity w' d is to be reduced 

to w b in a given length the accel- 
eration is given by 

w'd — w b w' d — w b w'd 2 — w b 2 



time length 2 length (19) 

w d + Wb 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 283 

If this occurs in a length 20d t , which is a length often used for 
the delivery tube, the following results : 

_ w'j - w b 2 
a - 404 (20) 

The velocity at any point at distance x from the throat is 
given by the equation 



But 



2x 



404 



204 L 1 \w'J J w'i 

UdA 



w 2 



Q . 

\rdl_J 



d A 

4 4 



(21) 



Hence 



1 '204L wvJ 4 4 

4 



do; = 



V 1 I 1 w' d V204 



Now in general w b will be equal to about 4 ft. per second and 

w/ 6 about 140 ft. per second so that —r^ = twf^- This is negli- 
v w' d 1200 

gible. Hence 



dx 



d t 



<l> 



(22) 



X 

204 



The solution of this is given by the following table. 



X 

57 


1 


2 


3 


4 


8 


12 


16 


19 


19.75 


d x 
dt 


1.013 


1.027 


1.040 


1.058 


1.138 


1.254 


1.496 


2.118 


3.00 



DENSITY 



The value of the density of a jet is found by obtaining the 
temperature of the discharge when the injector will just waste and 



284 HEAT ENGINEERING 

the weight of discharge in a given time. If then the theoretical 
amount in this time and with the pressure be divided into the 
actual amount of water handled, the square of this will give the 
density. If for instance an injector lifting 16,600 lbs. of water at 
148° F. will just discharge this quantity at 160 lbs. gauge pressure 
when the area at the throat is 0.00054 sq. ft. the density may be 
found thus: 

The weight of 1 cu. ft. of water at 148° F. is 61.22 lbs. Hence 
the head corresponding to 160 lbs. with a density of A is 

160 X 144 376.6 



61.22A A 

155.7 

a/a 

Weight discharged 



^ = 8.02^=.^ 



UK7 

16,600 = -^j=- X 0.00054 X 3600 X 61.22A 



IMPACT COEFFICIENT 

The value of k could be worked out after knowing A and find- 
ing the various velocities if z is determined experimentally. 
This has been done and the values found vary from 0.25 to 0.60. 

INJECTOR DETAILS 

The value of z is fixed by the various velocities and on account 
of the high velocity of steam acquired by discharges at even low 
pressures, exhaust steam could be used for feeding boilers or boiler 
steam could be used to force water into a chamber at a pressure 
much higher than boiler pressure. Thus an injector could be used 
for the hydrostatic test of a boiler. The injector shown in Fig. 131 
is a single jet injector although double jets are sometimes used. 
The first jet lifts the water and forces it into the second nozzle to 
be driven into the boiler. A lifting injector is one which will lift 
its supply and drive it while a non -lifting injector is one which will 
only force water. In injector testing the term maximum capacity 
means the greatest quantity of water which could be sent through 
the injector with a given steam pressure and temperature of feed, 
while the minimum capacity is the least that will be discharged 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 285 

without waste. The range of capacity is the difference of these 
expressed as a percentage of the maximum capacity. The over- 
flowing temperature is the highest temperature of operation with- 
out overflowing when working at a given pressure. The highest 
pressure obtainable without wasting is called the overflowing 
pressure. 

Injectors can work as pumps when operating with other fluids 
than steam, such as water. In such apparatus water under a 
great head acquires a velocity so high that it can lift several times 
its own quantity through a smaller height. The principle and 
equation of momentum would be the same for this apparatus. 
An injector operated by water is known as a jet pump or siphon. 
To apply the principles and equations above, suppose it is 
desired to have an injector pump 6000 lbs. of water per hour at 
68° F. into a boiler at 150.1 lbs. gauge pressure. The steam has a 
quality of 0.96. Assume no pressure on the suction and a lift 
of 6 ft. 

p t = 0.57 X 164.8 = 93.8 

w t = 223.7VH60.8 - 1116 = 1495 ft. per sec. 

w m = 223.7\/(1160.8 - 991.1)0.9 = 2760 ft. per sec. 

w c = 223.7\/(l 160.8 - 917.2)0.9 = 3310 ft. per sec. 

v t = 4.34 cu. ft. 

v m = 22.86 cu. ft. 

(for i = 991.1 + 17.0 = 1008.1 at 14.7 lbs.). 

w w = 8.02V3^[2.3 X ( - 4) + 34 - 6 = 20.1 ft. per sec. 



w n 



cno /(164.8- 4)144, 4 2 1lrK K ,. 
= 8 -° 2 V 60X0.8 + 64^ = 175 - 5 ft - Per SGC - 
1/2(3310 + 220.1) = (1 + 2)175.5 
_ 3310-351.0 
z ~ 351.0 - 20.1 " °' y0 

[1 20 1 2 i r 1 17^ ^ 2 ~\ 

36A + 778 X 6^4 J = 9 - 95 l/« + 778 X H4A J 

1160.8 + 323 + 0.072 = 9.95g' m + 6.17 

q' m = 148.5 

t, = 180° F. 
m m = 61.0 

The value of m m A of 60 X 0.8 will not have to be changed 
as this is as close as the value of A will warrant. It will be seen 



286 HEAT ENGINEERING 

that the expressions for the kinetic energies are so small that 
they may be omitted. 



:.02^< 



(164.8-14.7)144 4 2 ,„„„ 
61X0.8 --6Ti = 169 - 6 



Mass of steam per sec. = og . A w n . = 0.187 lbs. 

oolMJ X o.yo 

Mass of water in suction per sec. = 1.66 lbs. 

Mass of mixture per sec. = 1.847 lbs. 

F t = ° ,18 [ 4 q 5 4 ' 34 = 0.000543 sq. ft. 

_, 0.187 X 22.86 _ _ m _ ., 
F m = 2760 — = °* 00155 sc l- ft * 

F - = 169.6 X 61 7 X 0.8 = a000224 ^ ft ' 
F t _ 0.000543 
F d 0.000224 

This is between 2 and 3. 



/0.0002! 

V 67 



,0.000224 X 144 n „ . - 

d d = * /- ^o^r = °- 202 ln - = 5 mm * 



The size is therefore No. 5. 

d t = 0.314 in. 
d m = 0.60 in. 




Fig. 135. — -Shape for injector tubes. 

The table for the delivery tube is as follows: 

x = 0.202 0.404 0.606 0.808 1.616 2.424 3.232 3.838 

d x = 0.202 0.204 0.237 0.205 0.207 0.228 0.254 0.302 0.443 

The length of the delivery tube is 3.8 in. 

The length of the combining tube is 18 X 0.202 = 3.7 in. 

The length of the entrance to nozzle = 2J£ X 0.314 

= 0.78 in. 
The length of the diverging part of nozzle = 4 X 0.312 

= 1.25 in. 
This is shown in Fig. 135. 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 287 



STEAM TURBINES 

The steam turbine is operated by the force exerted when a 
steam jet strikes against moving blades. If a jet of a certain 
cross-section, discharging m lbs. of steam per second at a velocity 
of w a strikes a blade which is moving in the direction of the jet 
with a velocity Wb, the steam is reduced to this velocity in the 
direction of the jet. The acceleration of this substance is 



IV r 



w b 



At 



(24) 
During 



if it is assumed that this action takes place in At seconds. 

this time the amount of ^ 

mass acted upon is mAt. vJ 

For although all of the 

fluid leaving the jet 

would never strike the 

one vane considered, in 

all forms of apparatus 

there are other vanes 

which come into line so 

that the amount to be 

considered on the one 

vane is the mAt which 

strikes this vane (if desired At would depend on the number of 

vanes passing the jet per second). 

The force exerted due to this decrease of velocity is given by the 
general formula 

P = ma (25) 




Fig. 136. — Jet impinging on vanes. 



mAt 



w r 



w b 



At 



= m(w a — w b ) 



This expression is in absolute units of force, poundals in the 
English system or dynes in the French system. To change it to 
pounds or grams the expression must be divided by g giving 

m(w a — w p ) 



P = 
If w b is made zero this becomes 



P = 



mw a 



(26) 



(27) 



In this latter case there is no work done as the blade is sta- 



288 



HEAT ENGINEERING 



tionary. The force is called the force of impact or the impulse of 
the jet. The nozzle of the j et must feel a force equal to this as some 
force must be required to produce this flow. This is called the 
reaction of the jet. The word impulse refers to the force exerted 
by the jet when it strikes a body while reaction refers to the force 
on the body from which the jet issues in virtue of which the jet 
exists. Since in the case of the stationary blade there is no work 
done, there is no abstraction of energy and the velocity of the 
steam should not decrease. An investigation made on water jets 
showed this to be true. Although the velocity in the original 
direction is destroyed, the value of velocity of the water was not 
changed; its velocity in all directions along the plane being equal 
to the original velocity. 





Fig. 137. — Jet impinging on vane at Fig. _ 138. — Actual and relative 
angle with path of motion. velocities of the stream over a mov- 

ing blade. Entrance and exit tri- 
angles. 

In equation (26) the numerator represents the change of ve- 
locity in the direction of motion of the blade. If the plane of 
movement of the moving blade is not that of the nozzle so 
that the actual velocity w a is inclined at the angle a to the direc- 
tion of the motion of the blade Wb, equation (26) becomes 



m 



(W a COS a — Wb) 



(28) 



If in addition to this the substance is thrown off from the blade 
with an actual velocity w' n inclined at an angle a, then the change 
in velocity in the direction of motion as shown in Fig. 138, 
would be 



w a cos a — w' a cos a 



and hence the force would be 



P = —{w a cos a — w a cos a) 



(29) 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 289 

If there were no motion of the blade and no friction, w a would 
equal w' a and the force P would be the force exerted in the direc- 
tion marked w b . If however there is motion of the blade, w a 
does not equal w' a and there is a relation between these and the 
angles of the nozzles and blades if the best conditions prevail. 

If w a is the actual velocity from the jet in space and this is 
directed toward a blade moving with a velocity w b , the motion 
of the fluid relative to the blade w r is found as the component 
of w a by the triangle of velocities if w b is the other component. 
This means that to one standing on the blade and moving 
with it, the jet would appear to come in the direction w r at 
the angle (3 to the direction of motion; hence if the blade is to 
receive this stream, without impact or shock it should be tangent 
to this direction. If now the stream is conducted over the blade 
to the outlet side and sent off relative to the blade at an angle 
j8 r to the direction of motion and with a relative velocity w' r 
the actual velocity of discharge in space as the resultant of w' r 
and w b will be found to be w' a at the angle a. 

Of course it will be seen that as the blade moves away from the 
nozzle, the jet will not impinge on it but it must be remembered 
that another blade or vane will be moved up to take the place 
of the one shown in the figure. 

The work done by the force P is equal to the force multiplied 
by the distance moved through in the direction of P or 

Work = P X distance (30) 

Distance = w b X At (31) 

TYl 

Work = —(w a cos a — w f a cos a) iw b X At) (32) 

but mAt = M 

M 

.*. Workfor M lbs. = — (w a cos a — w' a cos a)w b (33) 

This is the work for M lb. in any time. If M is the amount per 
second this will give the work per second. 
The work may be separated into two parts : 

M -M , 

— w a cos aw b and — w a cos a w b 

and may be said to be equal to the work done at entrance by 
reducing the velocity to zero minus the work done at exit in 
giving the fluid its actual discharge velocity. 

19 



290 



HEAT ENGINEERING 



In other words the reaction of the discharge jet at exit mul- 
tiplied by the velocity of the blade subtracted from the impulse 
of the jet at entrance multiplied by the velocity of the blade at 
entrance is equal to the work per second. 

This method of statement in two parts is necessary when a 
turbine is built with radial flow in which w b is not the same at 
entrance and at exit as shown in Fig. 139. In this case 



M 



work = — [w b w a cos a — w' h w' a cos a] 



(34) 



In most cases steam turbines are of the axial flow type in 
which the steam flows across from inlet to outlet in a direction 




i Velocity of Whirl 
Entrance 




% ^f a 


i r 


\ t ,, \ 


\ w 6 


180 \B S 


\ix ' Wh 


\Zw' r 


\ w ' a 


/ -w b 


_il 


•_>j '^Velocity ofWhirl 
i I Exit 



Fig. 139. — Radial flow turbine. Veloc- 
ities Wb and w'b are different. 



Fig. 140. — Triangles of flow 
at entrance and discharge. 



parallel to the axis and not in a radial direction. Hence w b = 
w' b . Equation (33) shows that the work per pound of fluid is 



work per pound = - (w a cos a 



w' a cos a)w b (35) 



W b cos a and w' b cos a are the components of the actual 
velocity of the jets in the direction of motion and are called the 
velocities of whirl. Hence the work per pound is the difference 

w b • 
in the velocities of whirl at inlet and outlet multiplied by — 

It will be important to remember that if the velocity of whirl 
at exit from the blade is in the opposite direction to that at en- 
trance that this difference is really an arithmetic sum since the 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 291 

sign at exit is minus. This is shown clearly by the value of 
a. In all cases the angle is measured from w b when the arrows 
point toward or away from the vertex for w b and w a , or w b and w r . 
The graphical parts of Fig. 138 can be redrawn in Fig. 140. 

In this figure w r may be greater or less than w' r or equal to it. 
If there is no drop in pressure across the vane and if there is no 
friction 

W r = W' r 

If there is no drop in pressure and if there is friction w r is greater 
than w' r while if there is a drop in pressure there would be an 
increase in velocity due to the addition of heat energy. Thus 



w' r = ^2g[j(n - i 2 ) + |£|[1 - y] (36) 

This formula would give w' r if there is a drop of pressure and 
friction. Turbines in which there is a drop in pressure across 
the moving blade are called reaction turbines or pressure tur- 
bines while those in which there are no changes in the pressures 
as the steam passes over the moving blades are known as impulse 
turbines or velocity turbines. These names do not mean that 
impulse takes place in one form and reaction in the other. 
Reaction and impulse are present in each. These are only names 
borrowed from hydraulic turbines which have the meanings at- 
tached above. 

For impulse turbines with friction 

V)' r = W r \/l —y = fw r (37) 

/ = Vl^ . (37 r ) 

Now y depends on the velocity of the steam over the blades. 
The curves of Fig. 141 have been constructed from data given by 
Mover, for stationary and movable blades. S refers to the first 
and M refers to the second. 

In Fig. 140 there is no necessary relation between a, /? and /3' 
although a is fixed by the velocities and the three angles. The 
following trigonometric relations must hold between the triangles 
of entrance and exit: 

tan p = "• <** <* (38) 

WaCOSa: — Wb 



292 



HEAT ENGINEERING 



W r = 



w a sm a 



sin 



w b = w a cos a — w r COS (3 

W' r = W r -\/l — y = fw r 

W' a = \^w' r 2 + Wb 2 + 2WbW' r COS /3' 



sin q:' = 



w'r sin /5 r 



^ a 



(39) 

(40) 
(41) 
(42) 

(43) 

W' r COS /3 r = W r a COS c/ — W 6 (40') 

Thus if w a , a, and j8' are given in a problem the above 
equations must be satisfied if there is to be no shock and the 
equations give the values of the other quantities. A graphical 
solution is close enough in most problems and the construction 

0.40 



0.30 



■s 

S0.20 

I 



0.10 





























































s/ 


























M 













































































































500 



1000 1500 2000 

Velocity Relati-ve to Blade 



2500 



Fig. 141. — Values of y from data given by Mover. 

of the triangles of Fig. 140 will give the results of the above 
equations. 

To save space the lower triangle of Fig. 140 is turned through 
180° and placed so that its apex agrees with that of the upper 
triangle giving Fig. 142. In many turbines /3 and /3' are made 
supplements of each other giving Fig. 143. 

In this figure a represents the velocity of whirl at entrance, b 
represents in a reversed direction the velocity of whirl at exit 
and c represents the velocity of the blade. Hence to the scale of 
the figure, the work per pound of steam is 

c(a + 6) 



a [a~ (~b)] 
y 



(44) 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 293 



(45) 



The energy in the steam, as it strikes the blade, is 

2g 

since d represents the velocity of the jet leaving the nozzle. The 
efficiency of application of this jet to the blade is spoken of as 
the kinetic efficiency and is equal to 

2c(a + b) 



T)k 



cP 



(46) 



In this expression the scale of the figure need not be known 
as this is a relation between the lengths. 



A\ 





Fig. 142. — Diagram for in- 
let and outlet triangles with 
vertices at same point. Fric- 
tion on blade. 



Fig. 143. — Triangles with /3' the sup- 
plement of /3. Friction on blade. 



MAXIMUM EFFICIENCY 

If w b or c of Fig. 143 is increased, the quantity a would re- 
main the same although b would be decreased, so that it might 
increase the product if c were increased. In any event there 
would be a change and there must be a velocity which would 
give the maximum work. To find this there are two cases to 
consider, first if a is fixed with /5 = 180 — /3', and second if 
/3 and /3' are fixed. There is no need of considering the actual 
value of w a as this is only a matter of scale. Of course it must 
be of fixed value in the discussion, whatever that value is. 



Wb, 



work = — [w a cos a — w' a cos a] 

W' a COS a = W' r COS $' + Wb 

work = — [w a cos a — (w' r cos /3' + w b )] 



(35) 
(40') 



cos /3 = — cos 



V) 



fw r 



(41) 



294 



HEAT ENGINEERING 



Hence 

where 
Now 

Hence 



work = — [w a cos a + fw r cos /3 — w b ] 

W r COS (3 = w a cos a — w b 
work = — [(1 +f){w a cos a — w b }] 



(40) 
(47) 



In this the only variable is w b , hence the work is a maximum 
for the value of w b given by equating the first derivative to zero. 



d work 1 + / 

= = [W a COS a 



dwi 



2w b ] 



w b = ybw a cos a 



(48) 




Fig. a. — No friction. Fig. b. — Friction. 

Fig. 144. — Triangles of discharge for maximum efficiency without friction 

and with friction. 

This means that the velocity of the blades should be one- 
half the velocity of whirl. If / = or there is no friction, this 
same fact is true. The best result occurs when the wheel is 
moving at one-half the velocity of whirl. 

If there is no friction this will cause the absolute velocity of 
outflow to be perpendicular to the motion of the wheel, while 
if there is friction there will be a discharge at an angle to this 
perpendicular as is shown in Fig. 144, a and b. The work in this 
case becomes: 

work = — — ri 



9 



4 W a 2 COS 2 a — Ji Wa 2 COS 2 a] 



(i+/) 

4? 



w a 2 cos 2 a 



The kinetic efficiency is : 

work 

2? 



(1+/) 



COS z a 



(49) 



(50) 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 295 

If there is no friction / = 1 and this becomes : 

rj k = cos 2 a (51) 

The other case to be considered is that in which (3 and $' are 
fixed and a may be varied. 

Work = — (w a cos a — w' a cos a) (35) 

— — (w r cos |8 + w b — w' r cos ft — w b ) 

= — w r (cos (3 -fcos ft) (52) 

Now cos j8 — / cos j8' is a fixed quantity and w&w r are variable 
quantities. 

The efficiency in this case is: 

_ work _ 2w b w r (cos g — / cos ft') 
** = ~wJ~ ~ w? (53) 

2<7 

w a 2 = w b 2 + w 2 + 2w b w r cos /3 (42') 

2w b w r (cos |8 — / cos j8Q ' v 

** ~ m 2 + w 2 + 2w b Wr cos j8 { / 

or calling 2 (cos (3 — / cos /?') = /c and — L = R 



Vk — j 



R + o + 2 cos 






h (55) 

2 cos/3 



(i2 + | + 2 cos |8 ) 



# 2 = 1 
E = + 1 

.'. w r = w b (56) 

and the triangle at entrance is isosceles. 

= 2a 
Hence the work becomes: 

work = — (cos jS — / cos /?') 



296 




HEAT ENGINEERING 

w a sin a w a 


Now 


sin 2a 2 cos a 


Hence 




W a 2 (COS (3 — / COS jS') 

work - 1 , a 

2g 1 + cos /3 


Since 
for 




4 cos 2 a= 2(1 + cos/3) 

(cos /3 - / cos /3') 
^ " 1 + cos /3 

= 180 - /3' and / = 1 




cos/3 


_ COS _ 1 . w 1.//Q = 



(57) 



^2P = 1 —tan 2 a (58) 
3^(1+ cos jS) cos 2 H /3 

One other problem of maximum efficiency should be con- 
sidered. Suppose that it is desired to operate a given blade 
at a velocity w b , what should be the angle a and the velocity 

w a to give the greatest efficiency? 
From (52) : 

work = ^- r (cos - / cos /3') (52) 

In this w b , /3, / and (3' are known and hence the maximum 
work would occur when w r is as large as it can be made. Since 
residual energy w' a also increases, the efficiency may not be in- 
creased. This problem therefore depends on efficiency. 

work 2w b w r (cos/3 — /cos/3') 7 w r 
Vk = 



Now 



Wa 1 W a 2 


* Wa 2 


(59) 


ig 






w a 2 = w b 2 + w r 2 + 2w b w r cos /3 




(420 


kw r 
' Vk — ... o , ... o , r» _ a 




(590 



This is a maximum when 
j 

-. — = 0; or k(w b 2 + w r 2 + 2w 6 w r cos 0) = k(2w r 2 + 2t0&w r cos /3 

w r 2 = w 6 2 (60) 

or the triangle at entrance is isosceles. 
This fixes w a by the formula above 

w a 2 = 2w b 2 (1 + cos /3 ), 

w b (l + cos jS) , , 

or w a = — w^— (61) 

cos }i jS 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 297 

This is in reality the same proof as the former one since if 
for a fixed w b and blade the triangle is isosceles giving w a , 
for a fixed w a and blade the isosceles triangle should be the most 
efficient. 

The three cases have proven that where the angle a is fixed 
the turbine blade must travel at one -half the velocity of whirl, 
whether there be friction or not, and that if the angles /3 and ft 
are fixed the best efficiency with either w a fixed or w b fixed is 
obtained when the inflow triangle is isosceles. This holds 
whether there be friction or not. In the first case without fric- 
tion, the outflow will be perpendicular to the movement of the 
blade while in all other cases the line may be inclined slightly. 
If, however, the speed of the blade is desired and the angle a is 
fixed, then the highest efficiency is obtained if the velocity of 
whirl is made twice the speed of the blade. This fixes w a . 




Fig. 145. — Velocity compounding. 



In the above figures it has been seen that when the steam is 
used on one blade the velocity of this blade has to be equal to 
one-half the velocity of whirl if a is fixed and /3 and 180 — ft are 
fixed. This means that the velocity of the wheel would be very 
high. To reduce this, the velocity of whirl could be decreased 
by decreasing the velocity from the nozzle. This is obtained by a 
small drop in pressure in the nozzle. 

If steam is to be used through a large difference in pressure 
this would have to be utilized in a series of nozzles and blades. 
This is called pressure compounding. Another method is to 
use a series of movable and fixed blades as shown in Fig. 145, 
utilizing the high kinetic energy of discharge from one blade in 
successive blades. This is known as velocity compounding. 



298 



HEAT ENGINEERING 



The velocity diagrams are combined in Fig. 146, in which 
= 180 - ff 
0i = 180 - /3\ 
In this the work per pound is: 



work = - a + 6 + ai + 6i 



A\ 




Fig. 146. — Combined diagram for velocities in a turbine with velocity 

compounding. 



The efficiency is: 



work 2c(a + b + a\ + 6i) 




(62) 



Fig. 147. — Velocity compounding to give the best efficiency. 

The value of w b to make this or the work a maximum is desired. 

It has been shown that the last stage to give a maximum re- 
sult should have a velocity of blade equal to one-half the velocity 
of whirl and hence the figure to give the best result should be in the 
form shown in Fig. 147, in which c is one-half of the velocity of whirl 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 299 

in the last stage. To find the value of c to properly fit value of 
a of the diagram, continue the lines until they intersect the lower 

c " 

line. The first intercept is c. The second one is 7, since the 

slanting line w r has been decreased to w' r = fw r . From similar 
triangles 

lower intercept __ Wr_ 
c ~ fw r 



or 



intercept = 



7 



The second line is decreased in the same manner 

Hence the third intercept is (^ j or Kj Kj • 

The fourth intercept is the same as the third. Hence 

r 1 2i 

a = C L 1 + / + 7 2 J 



(63) 



2 5 6 




Fig. 148. — Construction for determination of c for multistaying with 

friction. 



If there were three velocity stages this expression would be 

.] (64) 



= c [ 1 + / + ? + /W 4 



To construct this so as to find c graphically: lay off any dis- 
tance 1—2, Fig. 148, which is called c; at right angles lay off the 
distance 1—3 equal to unity to any scale and 3—4 equal to /to 
this scale. Draw an indefinite line from 4 parallel to 1—2, pro- 



300 



HE A T- ENGINEERING 



ject 2 on this line at 2' and draw 3-2' to 5. 1-5 is the distance 

c c 

y Project 5 to 5', draw 3 — 5' to 6 and 1—6 is equal to j 2 - 



Hence 



[(1 - 2) + (1 - 5) + (1 - 6 ) + (1 



6)] = 
cfl+J + J] 

These are laid off from 3 to 6 "'and line 3— 7 is laid off equal 
to a, the velocity of whirl. If 7 and the last 6'" are joined and 
2" — 8 is drawn parallel to §'" — 7 the distance 3 — 8 will be 
equal to the correct distance c for Fig. 146. This same construc- 
tion can be used for any number of stages. 

To make use of the same angle of inlet into the various steps, 
the angle a of the second fixed blade is changed from the value 




Fig. 149. — Diagram in which a\ 



Friction on blades. 



180 — el to a. This is shown by the diagram in Fig. 149 where 
the line has been swung up to the original direction. In this way 
a\ is made the same as a and by this action a\ has been made 
larger than before and the efficiency has been increased. The 
various kinetic efficiencies for these arrangements of blades 
have been computed and result in the following: 



Fig. 144a 

Fig. 1446 

Fig. 147 

Fig. 149 



7j k = 81 per cent. 
T] k = 78 per cent. 
rjk = 73 per cent. 
7] Jc = 84 per cent. 



These have been drawn with a = cos ~ 1 0.9. 

The axial thrust is due to the difference between the impulse 
at entrance to the moving blade and the reaction at outlet in 
the direction of the axis. The force per pound of steam per 
second is: 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 301 



P = - [w a sin a — w' a sin a]. 



(65) 



Without friction this difference is equal to zero as is seen in 
Fig. 144. If, however, there is friction this formula does not 
reduce to zero but to a positive quantity. 

In the case of reaction turbines there is a static difference of 
pressure between the two sides of the blades which means an 
axial pressure. 

The actual forms of turbines will now be examined. The 
simplest turbine is the DeLaval turbine. 





Fig. 150. Fig. 151. 

Fig. 150. — Section through DeLaval rotor. 
Fig. 151. — Sections of DeLaval turbine with curves of pressure and 

velocity. 

Fig. 150 shows a section of the DeLaval turbine. This is an 
impulse turbine in which velocity is generated in a single set of 
nozzles A attached to a steam chest. The velocity is utilized 
on one set of blades. To increase the power of the machine a 
number of nozzles are used. The angle of the nozzle relative to 
the plane of the blade is made as small as possible, as shown in 
Fig. 150, so that the efficiency which is proportional to cos 2 a 
is as large as possible. The peripheral speed of the wheel is very 



302 



HEAT ENGINEERING 



great. w a is equal to 3820 ft. per second if the drop in the nozzle 
is from 150.1 lbs. gauge to 6 lbs. absolute. The speed of the wheel 
for a value of / of 0.95 and cos a = 0.9 would be 

Wh=z y 2X o.9 X 3820 = 1719 ft. per sec. 

This would mean 16,400 r.p.m. for a radius to the blades of 1 ft. 
The pressure drop for the axial distances of nozzle and blade and 
the absolute velocity changes are shown in Fig. 151. This figure 
gives a section through the axis and one parallel to the axis through 
the blades and nozzle. 

The Curtis turbine is shown in Fig. 152. In this steam enters 
the nozzle at A and is discharged against moving vanes. The 




Fig. 152. — Section through horizontal Curtis turbine. 

discharge from these moving vanes is guided by stationary vanes 
to another set of movable vanes, and after discharge from these 
it is taken to another set of nozzles B and discharges into a second 
set of vanes. In the figure shown there are five sets of nozzles, 
A, B, C, D, and E, and to each of these there are two movable 
and one fixed set of vanes. The pressure drop takes place in 
five stages, and there is no drop in pressure over the blades. 
The exhaust space F is connected to the condenser. This action 
is better shown in Fig. 153 in which a section through the axis 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 303 

and one parallel to the axis of a two-stage turbine are shown side 
by side following the method used by Moyer. In the figure the 
pressure is seen to be constant across the sets of vanes, the drop 
in pressure and consequently the gain in velocity taking place 
in the nozzles. The turbine is therefore of the impulse type. 
In some cases the nozzles which are arranged in sets extending 



Pressure 




Section 
Perpendicular 

to Radius 
through 

Blades 



Fig. 153. — Sections of Curtis turbine with curves of pressure and velocity. 

over a portion of the circumference are separated into two sets 
on diametrically opposite parts of the circumference so as to 
balance the forces in an axial direction. The axial thrust must 
be balanced by some form of thrust bearing G. An auxiliary 
valve is sometimes used to admit live steam into a lower stage 
under heavy loads. In this turbine holes are made through the 



304 



HEAT ENGINEERING 



discs to insure the same pressure throughout the moving elements 
of one stage. 

For the Rateau turbine the diagrammatic arrangement of parts 
is shown in Fig. 154. In this turbine there is only one set of 
blades to each set of nozzles and there is a drop of pressure in 
each fixed member. There is no change of pressure across the 




Axial Length 




Section 

Perpendicular 

to Radius 

through 

Blades 



Fig. 154. — Sections of Rateau turbine with curves of pressure and velocity. 

movable blades so that this turbine is also of the impulse type. 
As will be seen later the area through which the steam passes 
as the pressure falls must increase since the velocity relative 
to the blades is about the same in each set but the specific 
volume increases. Hence the length of the blades increases. 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 305 

In all of these the steam is admitted to a portion of the circum- 
ference and as the steam pressure falls the portion of the cir- 
cumference covered is greater so as to give greater area for the 
steam. The position of the successive nozzles will be advanced 
in the direction. of flow due to the advance of the steam as it 
passes over the moving blades. This is called lead. 

The Parsons turbine is shown in Fig. 155 and diagrammatically 
in Fig. 156. In this steam is admitted around the complete 
circumference to a set of fixed blades. These discharge on the 
movable set and these in turn to a fixed set from which the action 
is repeated. In each of these sets of fixed or movable blades 
there is a pressure drop on all blades and hence this type will be 
called reaction type of turbine. 




Fig. 155. — Parsons turbine. 



In this turbine the axial thrust is balanced by connecting 
balancing drums of proper area to the proper parts of the tur- 
bine and connecting the back of the last drum -to the space lead- 
ing to the condenser. A thrust bearing is used to ensure align- 
ment. In the type of turbine shown in Fig. 155 a double flow 
arrangement through the center A balances most of the thrust. 
To make the turbine more efficient the Curtis element C has been 
used for the first reception of the steam. The high-pressure 
steam is discharged from nozzles attached to the steam chest 
B. The steam from the movable blades C then passes to the 
fixed and movable blades at D and finally is discharged at E. 
At this point the steam divides into two parts, one to blades / 
and the condenser connection F and the other enters the space 
A at the center of the drum A through holes and thence to holes 
to the space from which the steam enters a set of blades and 
finally issues into the condenser from the space H. This rep- 

20 



306 



HEAT ENGINEERING 



resents one of the more recent forms of Westinghouse Parsons 
turbine. The blades are mounted on drums A, I and J of 
different sizes to allow for the changes in volume of the steam. 
The main steam valve of the ordinary form of Parsons turbine 
admits steam in puffs controlled by the governor; while under 
very heavy load the governor begins to operate on a second 




Absolute 
Telocity 



Axial Length 




i 



y 



la 



i 1 



zJ \ jj Section 

\ J, Perpendicular 

^ jj to Ka'dius 

^ jf through 

N Blades 



Fig. 156. — Sections of Parsons turbine with curves of pressure and velocity. 

valve admitting extra steam to the second stage of the turbine. 
The end thrust, although reduced by the two turbines placed on 
the same shaft in which the steam passes to right and left, may 
exist and for that reason a small thrust bearing is used to ensure 
alignment. The air leakage around the shaft into the steam 
space which is at the pressure of the condenser is prevented by 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 307 

some form of steam stuffing box. Fig. 156 shows how the pres- 
sure drops on both movable and fixed blades although the 
absolute velocity decreases over the movable blade. The 
decrease of absolute velocity would have been greater on the 
movable blade were there no drop of pressure here. 

The great disadvantage of the DeLaval turbine is the high 
blade speed and the resulting complications. For structural 
reasons the maximum output of this type is limited to about 
300 kw. The absence of packing and the simplicity of the 
machine do not make up for this great disadvantage. Between 
the reaction and the other impulse turbines there are advantages 
on both sides. The reaction turbines are very inefficient on the 
high-pressure stages, due to the leakage at the ends of the short 
blades; while in the impulse turbine packing around the shaft 
between the discs separating the stages is necessary. The com- 
bination turbine used by the Westinghouse Company in their 
use of a Curtis impulse stage for the first utilization of the steam 
and then the use of the reaction blading after a reduction of 
pressure has been made to combine the advantages of each type. 

EFFICIENCY 

The various efficiencies of the turbine may be computed. 
The nozzle turns the heat (i\ — 1*2) (1 — y) into kinetic energy 
and z'i — q' heat units have been required per pound. Hence 
the overall nozzle efficiency is 

_ Q'i - i 2 )(l - y) 
Vn ~ ii - q'o 

In some cases there is a series of nozzles in a turbine and 
if Qh #2, #3, etc., are the amounts of heat turned into work the 
average nozzle efficiency would be 

qx + #2 + #3 + • . • 

Vn = : -, 

ii - q o 

The kinetic efficiency is best worked out by a series of diagrams 
and is equal to 

_ 2c(a + b + a' + b' + . . .) 
*»-- -52- 

These quantities may be found graphically or computed analy- 
tically after drawing such a figure as Fig. 147. The use of 



308 HEAT ENGINEERING 

graphical diagrams for the simplification of analytical work in 
pointing out relations should be well understood. 

In the analysis of reaction turbines these two efficiencies 
cannot be separated. These are computed together as will be 
shown in the computation for a Parsons turbine. 

The loss due to radiation from the turbine may amount 
to 1 per cent, of the heat supplied and the loss due to leakage 
varies from 1 per cent, in the case of impulse wheels of the De- 
Laval form to 5 per cent, in the case of reaction turbines. The 
efficiency of transmission due to leakage and radiation may be 
taken together and called the efficiency of weight. 

t\ w = 1 — (loss of radiation + loss of leakage). 

The next item which reduces the delivery from the shaft of a 
turbine is the friction of the blades and discs which is known as 
windage. This amounts to about 5 per cent, for the DeLaval 
turbines, while for many other forms of discs it may amount to 
10 per cent. This depends on the pressure of the steam against 
the discs and is computed for any given problem by formulae 
given in text books on turbine design. In addition to this loss 
there is loss due to friction at the bearings which may amount 
to 1 per cent. In some turbines there are oil pumps and gears 
which consume 1 or 2 per cent, of the power exerted on the blades. 
If the sum of these be subtracted from unity the difference may 
be called the mechanical efficiency, or 

r) m = 1 — (windage loss + bearing loss + gear or pump loss) . 

The electric generator will have an efficiency of rj e . 
The overall thermal efficiency is therefore 

Vt = Vn X 7) k X y]w X f]m X f\e 

If now the probable steam consumption per kilowatt hour out- 
put is desired, the quantity is found as follows: 

One kilowatt hour = n nAa = 3410 B.t.u. 
0.746 

Amount of heat supplied per lb. steam = (i — q'o)- 

Amount of heat utilized per lb. steam = i\ t {i — q'o). 

3410 
Amount of steam per kilowatt hour = —r. r\ • 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 309 

SIMILARITY OF ACTION OF TURBINE AND ENGINE 

0*1-2 2 )(1 - #) 



The expression for the nozzle efficiency is 

2 Q o 

This is the expression for the efficiency of the Clausius cycle 
with complete expansion except for the friction term (1 — y). 
Moreover, the area on the pv plane representing the gain of kinetic 
energy is the area behind the adiabatic. This is the area of an 
indicator card with complete expansion. The theoretic action 
of the steam is therefore the same in the engine and in the tur- 
bine. In the turbine, however, the cool portions never come in 
contact with the hotter steam and hence there is no initial con- 
densation. For this reason there is no need of free expansion 
and the toe of the expansion line is utilized. Hence high vacuua 
are of value for turbines. The gain in the T.-S. diagram extends 
over the whole width of the figure in place of the partial width 
available in the engine with free expansion. 

In the computations for these machines the various types will 
be considered. 

COMPUTATIONS AND DESIGN 

DeLaval Turbine. — It is required to get the leading dimen- 
sions of a DeLaval turbine to develop 200 kw. with dry steam 
at 132.5 lbs. gauge pressure and the back pressure is 2.5 lbs. 
gauge. 

Vi = 147.2 lbs. abs. ti = 1192.2 
p t = 83.8 lbs. abs. i t = 1146.7 
p m = 17.2 lbs. abs. i m = 1034.8 

q' m = 188.4 

w m = 223.7VVi - 1*2) (1 - y) = 223.7 y/{\ 192.2 - 1034.8)0.88 

= 2630 ft. per sec. 

w t = 223.7VVi - it = 223.7VH92.2 - 1146.7 = 1508 ft. per sec. 
v t = 5.043 cu. ft. 

v m = 20.75 cu. ft. (for p = 17.19 and i = 1034.8 + 0.12 X 157.4) 
Assume a = 20° 

w b = y 2 X 2630 X cos 20° = 1315 X 0.9397 = 1235 ft. per sec. 

. a 2630 X 0.342 _ ^ 00 
tan (3 = --^z. = 0.728 



310 HEAT ENGINEERING 

= 36° - 2' = 36° 

2630 X 0.342 ■ 
w ' = ~^588 = 153 ° 

/ for 1500 ft. per sec. = 0.9 

1 Q0 
Kinetic efficiency = -y- X (0.94) 2 = 0.840 

i ax- r i (1192.2-1034.8)0.88 „, 000 
Thermal efficiency of nozzle = - — 1ino Q , = 0.1382 

Radiation and leakage = 1 per cent. 

5 per cent, for windage ] 

Friction loss 1 per cent, for bearings j = 8 per cent. 

2 per cent, for gears j 

Generator efficiency = 95 per cent. 

Overall efficiency = 0.1382 X 0.99 X 0.840 X 0.92 X 0.95 

0.1008 
= 10.08 per cent. 

2546 
746 
Steam per kw.-hr. = (1192 . 2 _'i 88 .4)0.1008 = 33 ' 8 lbs * 

Total steam per hour = 200 X 33.8 = 6760 lbs. 

Lbs. per sec. = 1.87 lbs. 

n w a t 1 xl + 1.87X5.043X144 
Combined area of nozzle throats = TzFiQ = 

0.903 sq. in. 

n w a * +u 1.87X20.75X144 01 _. . 
Combined area of mouths = ofvx) = 2.124 sq.m. 

If 10 nozzles are used the areas of the throat will be 0.0903 
sq. in. and at the mouth 0.2124 sq. in. The diameter will be 
0.339 in. and 0.520 in. respectively. The nozzle will be cut away 
on an angle so that its length along the face of the blade will be 

0.520 X -t 1 — = 0.520 X t^jt, = 1.52 in. 
sin a 0.342 

Let the blades be made as shown in Fig. 157. The thick 
part in the center being introduced to stiffen the blade and also 
to keep the area through the blades of a constant width. The 
area required for the blade passage per nozzle is 

1.87X20.75X144 



8 X 1530 



= 0.457 sq. in. 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 311 



The normal entrance equals this amount and the area along 
the sides of the blades will be 



0.457 0.457 



= 0.778 



sin jS 0.588 

If now the length of the nozzle is 1.52 in. the 
height required to take the steam will be 

0.778 



1 .52 — blade thickness 



= 0.50 in. 



This is about the height of the nozzle and hence 
if the blades are made 1 in. high there will be little 
chance for leakage. 

In order to have space for 10 nozzles suppose 
the radius of the disc be taken as 18 in. Then the 
number of revolutions for this turbine would be 




N 



1235 X 60 



7850 



2tt X 1.5 
If this is desired at 8000 r.p.m. 

1235 X 60 



Fig. 157. 
— Arrange- 
ment of 
blade thick- 
ness to give 
uniform 
p a s s a g e 
width. 



radius 



2tt X 8000 



X 12 = 17.65 in. 




Fig. 158. — Velocity diagram for a two-stage Curtis unit with varying 
values of /. 

Curtis Turbine. — It is desired to find the leading dimensions 
for a Curtis turbine with four pressure stages between 175 lbs. 
gauge pressure and 125° F. superheat and 2 in. absolute pressure 



312 HEAT ENGINEERING 

at exhaust, each stage to have a double velocity stage or two 
moving blades and it is required to have w a from nozzles on each 
stage the same and the velocities w b the same on each disc. 
The power to be developed is 5000 kw. 

In such a case it is well to begin by drawing the diagram for 
velocities and from this compute the kinetic efficiency, as this 
must be known before the pressures of the various stages may be 
found. The value of / will not be the same for the various stages, 
since the velocity over the first movable blades is about 1500 
ft. per second, over the fixed blade is about 1200 ft. per second, 
and over the second movable blade is about 700 ft. per second. 
The values of / will be 0.9, 0.87 and 0.95. These varying values 
of / make the construction of Fig. 148 a little more complex as / 
is not the same on each. Assuming a = 20° and constructing 
the various values of / for Fig. 148, Fig. 158 has been con- 
structed. From this 

Ma + 6 + a' + ftp n799 
^2 = 0.722 = rj k 

This means that 1 — rj k or 27.8 per cent, of the kinetic energy 

of the jet of steam remains in the steam when it leaves the vane. 

w' r 2 
This is partially in the form of kinetic energy equal to -^~ and 

part in additional heat content from the friction. This heat is 
therefore to be added to the isoentropic heat content i 2 at dis- 
charge from the nozzle to find the heat content at entrance into 
the next nozzle. Whether the kinetic energy is considered as 
heat or kinetic energy the effect is the same as shown by equation 
(4) of this chapter. 

REHEAT FACTOR 

The effect of the heat is to increase the quantity available for 
the lower stages. This is shown clearly on the Mollier chart. 
Steam flowing from condition 1 to condition 2 should have the 
heat content i\ — i 2 changed into kinetic energy. On account 
of friction in the nozzle only (1 — y)(i\ — i 2 ) is changed into 
kinetic energy and in the Curtis turbine y may be taken as 0.06. 
This means that 0.06 (z'i — i 2 ) remains as heat in addition to i 2 . 
On account of the kinetic efficiency of the blades the heat (1 — -q k ) 
(1 — y)(ii — i 2 ) still remains in, in addition to the amount which 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 313 



should remain, 
stage is 



Hence the heat content at outflow to next 



iout = i2 + yi(ii — 12) + (l - 
= *i — yk(l — y)(ii — ii) 



Vk)(l - y)(ii - 12) 



(64) 



This last statement is equivalent to saying that the heat con- 
tent remaining is equal to the original heat content minus the 
work, which gives the point 3 for the proper heat content at the 
pressure of 2, hence the condition of outflow from the first stage 
or inflow to the second stage is given by the point V at which the 
pressure is the same as at 2, but i is the i of 3. The entropy s 




Entropy -s 

Fig. 159. — Mollier chart arranged for investigation of reheat factor. 

has been increased. If 1' — 2' is made equal to 1 — 2 and the 
same friction is assumed, the velocity of discharge will be the 
same as that for the first stage and the reheat, as the distance 
3 — 2 is called, will be the same since the same velocity diagram 
is used for this stage, so that the kinetic efficiency will be the 
same. This can be continued to l" - 2" and l 7 " - 2'" . 
Since the distances 1 - 2, l' - 2', l" - 2" and l'" - 2'" 
are all the same the velocities from all nozzles will be equal and 
the diagram of Fig. 158 is the same for all stages. Hence the 
angles of similar blades will be equal as will the velocities of 



314 HEAT ENGINEERING 

the blades. If the various pressure lines are continued back to 
the entropy line through 1 and 2 and the intercepts marked 
a, b and c, it will be found that on account of the form of the 
diagram, spreading out as entropy increases due to the 
properties of steam, 1 — 2 is greater than 2 — a; 2 — a is greater 
than a — b ; and a — b is greater than b — c. In other words, 5 
times 1 — 2, which is the heat available, is greater than 1 — d, 
the heat which would have been available for a single expan- 
sion. The ratio of the length 1 — 2 to the length — ^ — or 

heat on adiabatic for one stage . „ . , . 

is called the reheat factor. 

n 

The determination of this has been discussed by Edgar 

Buckingham of the U. S. Bureau of Standards, and is found in 

Vol. vii of the Bulletin of the Bureau of Standards, Washington, 

D. C. It is also given in Reprint No. 167. If this factor is called 

(R. H.) and is known in any case, it is immediately possible to 

find the length 1—2 for all stages since 

1 - 2 = f-=i) (R. H.) (65) 

Since 1 — 3 of Fig. 159 is the heat actually turned into work 
and 1 — 2 is the heat applied, the efficiency of each stage is 

_ 1 ~ 3 

This is the product of 77 n X -q k or ?7 S . 

In the triangle 321' in the saturated region 

is — ii = 3 — 2 = (xv — x 2 )r 

since the change in heat along a constant pressure line is equal 
to the change in quality multiplied by the heat of vaporization. 

T 

Now s v — s 3 = 1' — 3 = (xv — X 2 ) 7p 

2-3 (ay - x 2 )r 
Hence 1^3 0r tan 8 = ~ T = T 

(XV - Xz)^ 

or the slope of the curve of constant pressure with the hori- 
zontal is proportional to the temperature or 

tan 8 = kT (66) 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 315 

Since T is constant for a given pressure in the saturated region 
5 is constant and these lines of constant pressure are straight lines 
and since i = when s = they should pass through the 
origin although this is not strictly true for after condensation 
sets in the variation between i and s is logarithmic in form. 
However the divergence between these lines is so slight that the 
slope will not change much if they are assumed to pass through 
zero. Buckingham calls attention to the fact that a slight 
change in the lines will bring this about. With this understand- 
ing as to the origin 

tan 8 = - (67) 

s 

If now 7 be the angle between 1 — 1' and 3 — 1' the tangent 
may be written as 

di 

tany = - j s (68) 

The negative sign must be used since di is negative when ds 
is positive during expansion. 
It is possible to write 

1 - 3 





udii 7 — — j/ _ g 


Now 


1 - 3 = n, 1 - 2 


and 


v 3 2-3 (1 - „.) 1 - 2 

tan 5 tan 8 




Vs Vs i 




. . tan t — -, tan 8 — ., 

1 — Vs I — rj s s 




di rjs i 

ds 1 — T} s s 




di rjs ds 



Hence 



or 

l 1 — rj s S 

If this is integrated from 1 to the last point say l v the following 
results : 

T i v rjs , Si 

Log* - = ^7; io ge - 



Since 



(69) 

U _ Sd 
^ v ~ s v 



316 HEAT ENGINEERING 

id i y 
because tan 8 = — = — 

Sd s v 

but Sd = Si 

. Si id 

' ' s v ~ F 

It is also seen that tan 5 = kT, hence 

i d = hT d s d . 

ix = IcTiSi = hTiSd 



and 

Substituting this above 



T 

'. id = i\ Tjr (70) 

id _ i\ Td 
i Y ~ i v Ti 



= (iiTd\ 



F_ (T d \* 
i\ 



i\ 



-••='.(■-©') ™ 



Now if the efficiency of the stages were all r) s , the total heat 
drop usable on one stage would be -q s {ix — id) and this would 
be equal to 

Since id = kT d Sd 

ix = kTxSx 
Sx = s d 

amount available with reheat 



Now the reheat factor 



amount available with no reheat 
'TdV 



m 



"« I 1 - TJ 
For any given problem in which the limiting pressures and 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 317 

stage efficiency are known the temperatures may be found and 
this formula may be computed. The temperatures to be used 
are those of saturation and in the superheated regions these 
lines of constant pressure bend a slight amount making a slight 
error in the application of the formula but not enough to prevent 
its use as a guide, at least if the saturation temperature is used 
for the upper temperature. The formula is true for an infinite 
number of differential steps while for a finite number of definite 
steps the reheat factor would be slightly different. The formula 
is therefore of value to fix the first value of 1 — 2 although this may 
have to be adjusted in computing a problem. 
The value of rj s for this problem is 

r) s = 0.94 X 0.722 = 0.679 
T of saturation for 175 lbs. = 370.5 + 460.7 = 831.2°. 
T of 0.98 lbs. abs. pressure = 101.1 + 460.7 = 561.8°. 

561 .8\ - 68 



/ 561.8 X 
X831.2/ 



R jj . V831.2/ 0.23 

n*o/i 561.8\ 0.68X0.324 1,U 
°' 68 i 1 " 83L2J 

The Mollier chart repared by Marks and Davis will be used 
for this problem: 



Vi 


= 189.7 


superheat = 125° F. Si = 1.646 i x 


= 1297 


Vc 


= 0.98 


quality = 0.82 s c = 1.646 i c 


= 918 




R.H. {l 


— ic 
4 


i nA 1297 ~ 918 no n 

■■ 1.04 . = 98.7 = % x — 

4 


12 


Reheat = (1 - 


- 0.68) X 98.7 = 31.6 B.t.u. 








ii = 


= 1297 - 98.7 = 1198.3 








iv = 


= 1198.3 + 31.6 = 1229.9 








iy - 


= 1229.9 - 98.7 = 1131.2 








iy = 


= 1131.2 + 31.6 = 1162.8 








iy> - 


= 1162.8 - 98.7 = 1064.1 






V2 = 


■ 75 


Superheat = 32° F. 






Pv = 


■■ 75 


s = 1.684 superheat = 95' 


D F. 




Vv = 


: 22 


x = 0.97 






Pi" = 


22 


s = 1.728 superheat 6° F. 






P2" = 


5.3 


x = 0.93 




i\"t - 


= 1065.9 + 31.6 


= 1095.7 pi« = 5.3 s = 1.781 


x = 0.965 



iv = 1095.5 - 98.7 = 996.8 p v » = 1.0 s = 1.781 x = 0.895 



318 HEAT ENGINEERING 

The final pressure desired was 0.98 lbs. and this result is as 
close as can be expected. The value of i\ — i 2 is correct and w a 
may now be found. 

w a = 223.7V98.7 X 0.94 = 2146 ft. per sec. 

This holds for each nozzle. 

The thermal efficiency of the nozzles and blades is therefore: 

_ rj s (n - i i)n _ .68 X 4 X 98.7 _ 
77 ~ H-q'o ~ " 1297 - 69.2 " °- 219 

The overall efficiency is found by assuming- 

Vw = l - (0.01 + 0.02) = 0.97 

7} m = 1 - (0.07 + 0.01) = 0.92 

Ve = 0.96 

rj t = 0.219 X 0.97 X 0.92 X 0.96 = 0.188 

= 18.8 per cent. 

3410 
The probable steam per kw.-hr. = ^ i as y 1 9 ~ 7 ~ = 14.8. 

The actual amount guaranteed by the builders was 15.75 
lbs. per kilowatt hour. 

Total steam for 5000 kw. = 5000 X 14.8 = 74,000 lbs. per 
hour. 

Steam per second = 20.55 lbs. 

The nozzles must be investigated for the presence of the throat. 

p tl = 0.57 X 189.7 = 108 

Vm = 0.57 X 75 = 42.8 

p a = 0.57 X 22 = 12.55 

p H = 0.57 X 5.3 = 3.02 

There is a throat in each nozzle and i at these pressures is 
found from the chart of Marks and Davis. 

i(for 108.0 lbs. s = 1.646) = 1230, 104° superheat = quality 

i(for 42.8 lbs. s = 1.684) = 1180, 42° superheat = quality 

i(for 12.55 lbs. s = 1.728) = 1118, 0.97 = quality 

i(foT 3.02 lbs. s = 1.781) = 1059, 0.94 = quality 

w t = 223.7\/l297 - 1230 = 1830 ft. per sec. 

«V = 223.7\/l229.9 - 1180 = 1580 ft. per sec. 

w t n = 223.7\/ll62.6 - 1118 = 1493 ft. per sec. 

w t >» = 223.7\/l095.7 - 1059 = 1350 ft. per sec. 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 319 
The values of the specific volumes are given below: 



v t = 


4.99 cu. ft. 


Vff = 


10.56 cu. ft. 


V t " = 


30.2 cu. ft. 


V t '" = 


110.0 cu. ft. 


v m = 


6.24 cu. ft. for p = 75, i = 1297 - 0.94 X 98. 


v m > = 


18.0 cu. ft. for p = 22, i = 1229.9 - 93 = 


V m " = 


64.8 cu. ft. for p = 5.3, i = 1162.2 - 93 = 


V m '" = 


307 cu. ft. for p = 0.98, i = 1095.5 - 93 = 



I = 1204 
1136.9 
1069.2 
1002.5 

The specific volumes of the steam leaving the blades is equal 
to the specific volume for the pressures at these points and for 
the heat content of exit plus the sum of the friction loss in the 
nozzles and on the blades. The friction loss of the blades plus 
the residual kinetic energy is equal to (I-77) or friction equals 



1 — 77 — r.e. 



where r.e. is equal to 



From Fig. 158 £ - (J 



0.065 



Friction loss on blades = 1 - 0.722 - 0.065 = 0.213 
Total friction effect = 0.06 + 0.94 X 0.213 = 0.260 
Heat from friction per stage = 98.7 X 0.260 = 24.7 B.t.u. 

Vo = 6.51 for p = 75, i = 1198.3 + 24.7 = 1223.0 

*V = 18.30 for p = 22, i = 1131.2 + 24.7 = 1155.9 

!V = 65.5 for p = 5.3, i = 1063.9 + 24.7 = 1088.6 

v '" = 303.4 for p = 0.98, i = 996.8 + 24.7 = 1021.5 

From Fig. 156 

1 54 
Wr' = ^ X 2146 = 690 ft. per sec. 

The areas then at the various points in square inches are given 
as follows: 

_ 20.55 X 4.99 X 144 . 

= 1830 = sq * m * 

20.55 X 10.56 X 144 
Ft ' = L580 = sq * m ' 

29.55 X 30.2 X 144 
Ft" = j^g3 = 59.70 sq. in. 



320 HEAT ENGINEERING 

20 .55 X 110.0 X 144 
*"' = 1350 = Sq * ln * 

20.55 X 6.24 X 144 
F m = 2146 = Sq * m * 

20.55 X 18.0 X 144 _. ^ 
F m ' = 2l46 = 2 * 70 sq * m ° 

20.55 X 64.8 X 144 
Fm" = 2146 = 8 °-0 sq. in. 

20.55 X 307.0 X 144 . oon . 
F m '" = 2l46 = 2 *° sq * in * 

The outlet area from the last blade of each stage is given by : 

20.55 X 6.51 X 144 
Fo = 690 = 27 ' 9 Sq " m - 

20.55 X 18.30 X 144 „ ft ' . 
Fc = gQQ = 79.7 sq. in. 

20.55 X 65.5 X 144 001 ft 
°" = 690 = sq * m * 

20.55 X 303.4 X 144 1Qnnn 
Fo'" = ™ = 1300.0 sq. m. 

For the outlet area from the blade of each stage use specific 
volumes at about one-third the difference between the volumes 
for and m, using from Fig. 156, 

3 49 

-r 1 ^ X 2146 = 1515 ft. per sec. 

4.96 r 



W r > = 



20.55X6.36X144 10/1 
n J515 = 12.4 sq.m. 

20.55 X 18.17 X 144 
Fo' = y^t? = 35.5 sq. in. 

20.55X65.3X144 1fVrfl 

F'o» = t^t^ = 127.6 sq. in. 

1515 

20.55 X 306.9 X 144 
F'o>» = j-^te = 600.0 sq. in. 

For the outlet from the fixed blades the velocities are each 
equal to 1040 ft. per second and the specific volumes may be taken 
as the mean of those in the last two cases, giving: 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 321 

F f = 18.4 sq. in. 

F f > = 52.0 sq. in. 

F r = 186.0 sq. in. 

Ff = 870.0 sq. in. 

If the nozzles of the first stage be made of J-£ sq. in. cross 
section of mouth there would be nine of these on each side of the 
turbine. These will give an area of 9 sq. in. where 8.58 have been 
required. If these are made % in. wide the length of each on 
the circular face will be: 

y 2 

ot — ; — TiKS = 1-95 m. 
% sin 20 

These will take up about 30 in. of the circumference if there is 
allowance made for the partitions between each two nozzles. 
The throats of these nozzles will each be: 

w 8.06 

X A X g^g = 0.47 sq. in. 

or 0.75 X 0.63 in. 

For the second stage the mouth may be made 1 sq. in. in 
area requiring 13 on each side. If made 1 in. wide the length will 
be 

ok = 2m in - 

19.7 
The throat will contain 26 X ^ = 20.8 sq. in. If this is 

1 in. wide the depth will be 

20.8 



26 X 0.34 



= 2.35 in. 



These will take up about 42 in. on each side. 

If the areas for the third stage are made 3 sq. in. each there 
will be sixteen necessary on each side and these may be worked 
out as before using 1% in. for the width. 

On the last stage the area of each might be made 9 sq. in. 
requiring 25 on each side and the width would be 4% in. 
These would take up about 160 in. on a side or 320 in. in the 
circumference. If this were to use the complete circumference 
the diameter of the turbine pitch circle would be 102 in. or 8 ft. 
6 in. If this is considered too large the width might be made 
6 in., this would require 92 in. diameter. The blades of the first 

21 



322 HEAT ENGINEERING 

movable set would be slightly higher than the width of the nozzle 
mouth. They will be determined by their area at discharge. 
These will have to be investigated from 

Fa = length X height X sin jS. 

Now F m = length X height X sin a. 

Hence since length is the same for nozzle and blades 

u . i , , F sin a 

helght ° " h "F^In7 

h ° - ^ x 8i x uf[ = °- 9 in - 

35.50 0.34 
h °' = 1X "247 X Oil = L19 m - 

1 27 6 *H 

h » = 1.75 X -~- X^j = 2.20 in. 

h*> =4^X §0 x jgf = 5.3in. 

The heights of the fixed blades will have to be made longer than 
the first movable blades because although the specific volume 
increases slightly, due to friction, there is a marked decrease in 
axial velocity as seen from Fig. 158. 

18 4 ^4 
*'" ^ X 8^8 X 0^54 = 1 - 015in - 

*"= ixg?x£g = 1.32in.. 

V = 1.75 X ^ X ^g = 2.3 in. 

870 24 
V' = 4Kx|^X™ = 5.83in. 

For the outflow edge of the second movable blade the heights 
are given by: 

/»'.- «xg|x||=1.07in. 

79.7 0.34 
°' = X 2^7 X 0~79 = m * 

h' „= 1.75 X ^ X ^|| = 2.41 in. 

1300 0.34 
h'o»> = ^A X ^22 X 0.79 = 5 ' 98 in> 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 323 

In this it is seen that the blades will increase from %-in. nozzle 
to 0.9-in. movable blade, 1.02-in. fixed blade and 1.07-in. second 
movable blade or, to allow for spreading, say % in., 1}£ in., 
134 in. and 1% in. These for the second stage would be 
1 in., 1% in., 1}^ in- and 1% in.; for the third stage 1% 
in., 2% in., 23^ in. and 2% in.; and for the last stage 4J£ 
in., h}/2 in., 6 in. and 634 in. 

The angles of the nozzles would be 20°, the first movable 
blade would have angles of 24° at entrance, the fixed blade would 
have an angle of 33° and the last movable blade would be 52°. 
The outlet angles are supplements of the above as the blades are 
all symmetrical. 

The speed of the wheel, from Fig. 158 is 

j-^77 X 2146 = 433 ft. per sec. 
4.96 r 

The speed with a diameter of 102 in. would be 975 r.p.m. 
If 1000 r.p.m. be taken then D = 100 in. This is a possibility. 
The customary speed is 1500 r.p.m. showing that a smaller 
diameter is used. To use this diameter of 66 in., the heights of 
the vanes would have to be increased by 50 per cent. This 
would make the last blades 9% in. long. 

Rateau Turbine. — In this turbine there are many pressure 
stages and only one velocity stage to each pressure stage. If 
this is the case and 20° is the value of a and the desired speed of 
blades is about 500 ft. per second, the velocity across the vanes 
will be 750 ft. per second which gives 

/ = 0.94 

1 _i_ Q4 

rik = 2 ' cos 2 a = 0.97 X (0.94) 2 = 0.857 

Vs = vn X t]k = 0.94 X 0.857 = 0.805 

Suppose the turbine is to work through the same range of 
pressures as the Curtis turbine above: 

/ 561.8 \ 
\831.27 

0.805 (l 



/ 561.8 \°- 805 



561. 8\ 0.805 X 0.324 

831 .2, 



The heat required to give a velocity of 500 ft. with a = 20 c 
and / = 0.94 is found thus: 



324 HEAT ENGINEERING 

2w b 1000 _ 

w a = = ^r^- = 1065 

cos a 0.94 
The available heat for this is 

/1 065 \ 2 
i x - i 2 = 0.94 (^37) = 21 - 5 B -t-u. 

The amount required per stage is 
21.5 



1.035 



= 20.4 



at u t 4 1297-918 loly 
Number of stages = sttt — = 1°*' 

Use 20 stages. 

ii - t. = 1297 2 q 918 X 1.035 = 19.7 



w = 223.7V19.7 X 0.94 = 961 ft. per sec. 

961 X 0.94 
w& = n = 452 ft. per sec. 

The heat returned at each stage = 19.7 X (1 - 0.805) = 3.84 
B.t.u. 

Heat used = 19.7 - 3.84 = 15.06 B.t.u. 

Actual thermal efficiency = 1 907 _ ftQ 9 = 0.245 

Overall efficiency = 0.245 X 0.97 X 0.90 X 0.96 = 0.205 
The steam consumption per kilowatt hour would be 

3410 i\, „ E ,u 

M = 55 66 X 1127.8 = 14 ' 75 lbs ' 

Parsons Turbine. — In this turbine the reheat factor will 
have to be worked out from the blade efficiency of one set of 
blades. In these blades there is a drop in pressure on each set 
of blades whether they are movable or stationary. If there is 
the same amount of heat added on each set and if the velocity 
at entrance is the same into each set there will be the same ve- 
locity of exit. If for instance, a from the first set of fixed blades 
is equal to 20° and /3 of the movable blade is made equal to 
75°, the value of w b would be equal to 

w a sin a r sin al 

w b = w a cos a - --^ = ».|cOfl a - ^J 

Hence for any desired value of w b , w a could be found. 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 325 



w r 



w a sin a 



Wr = fw r = f 



sm/3 

w a sin a 



sin/3 



(73) 
(74) 



If now it is desired to use an angle 180 — $' = a for the angle 
of discharge and have the absolute direction of the jet at the 
angle a' = 180 — fi = 105°, so that the same form of blades, 
although right and left handed, may be used for each stage, 
it will be necessary for i 

w' r actual = w a * 




Fig. 160. — Blades of a Parsons turbine. 
Hence wJ above must be increased to w a or 



w, 



,.^r(H-.«a-r) + .(/=^ 

(223.7) 2 



W n 



'} sin a\ 2 



/ J sm a \ 
\ sin B I 



(ii - *' 2 )(l - y) 



ll — 12 



sm/S 

-•[' - my] 



(1 - y)(223.7y 



(75) 



(76) 



Hence if a, )8, and Wb are assumed, w a and w r may be found 
and from w r , f may be selected, then i\ — ii for the movable 
blade. 



326 HEAT ENGINEERING 

For the fixed blade the steam enters with the velocity 
, Wr sin a w a sin a 

Wa = = T~ = : ^T = W r 

sin /3 sin j8 

It leaves this fixed blade with the velocity w a , the same as 
that of the first blade after being affected by friction and gaining 
kinetic energy from heat. 

'f sin «\ 2 



i i — i 2 



, -I'-e-s)] 



(1 - y) (223.7) 



This does not give quite the same value as the first expression 
on account of the difference in /. 

Now the work done on this stage per pound is 

Wh 

work = — [w a cos a — w' a cos a] (77) 

Wb r„ , sin a cos /3 

= — w a cos 



Jl + Sm " C ° Sg l (78) 

L sin p cos a J 



sin /3 cos 
The heat used is i\ — i a +'i-i — i'i = Q (79) 



. w b f tana"] 

Hence A -™ a cos «|_1 + ^J 



i\ — i* + *'i — ^2 



(80) 



For the angles mentioned above and Wb = 300 ft. per second : 
_ 300 _ J*00^ _ 

W<1 " n n, 0.34 " 0.849 " dM 

°' 94 - 373 

H4 

w r = 354 X ^ = 124 

/ = LOO 
. . / 354 X2 ( 1 -[ LQQ X^ ] 2 ) OQn 

* - *■ = to; ■ i - o.o4 ■ = 2 - 30 

/ for the fixed blade is the same as the value used above, hence 
i\ _ i\ = 2.30 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 327 

m , , 300 X 354 X 0.94 r , 0.36l 
Work per pound per stage = ^2~Tfi L 3~73j 

= 3400 ft.-lbs. 
= 4.38 B.t.u. 

4.38 
Stage efficiency = j^tj = 0-95 = rj a 

The reheat factor may now be applied in the usual manner 
for rj s = 0.95. This is, for the same pressures as before, 

R ' H - = 0.95 X 0.324 = L013 

Total heat available = (1297 - 918) X 1.013 = 384 B.t.u. 

Total number of stages if velocities are the same on each = 

384 

^7q = 83 stages. 

This answer is not given in correct form as it is customary to 
enlarge the drum as the pressure falls. If the first drum is of 
diameter D lf the second is often made 

D 2 = a/2 Di 

and Z> 3 = a/2 D 2 = 2D X 

The speed will be proportional to the diameters and hence if 
the above is assumed for the middle portion 

»,= ^=° = 212 

V2 

w hz = 300V2 = 424 

4.6 
(ii — i 2 ) for first stage is -~- = 2.3 

(ii — i 2 ) for last stage is 4.6 X 2 = 9.2 

379 
Stages in first portion = Q Q = 55. 

vStages in second portion = w . n = 27. 

3 X 4.6 

379 

Stages in third portion = Q n =14. 

Total number = 96 stages. This is rather high. A common 

rule for the total number of stages is 

at u r * 2400000 /01 , 

Number of stages = « — (81) 



328 HEAT ENGINEERING 

In case above: 



N first : /^.^o = 54 



■LV second {Qf\f\\2 — 



^V*A*rd : (AOA\2 ~ ^ 



2400000 

(212) : 
2400000 

(300) : 
2400000 

(424) : 

The other quantities may be computed as shown in previous 
problems. 

ALLOWANCE FOR CHANGE OF RUNNING CONDITIONS 

Where turbines are actually tested and one condition is changed 
without changing other quantities the following allowances are 
found to hold: 

From 0°-100° superheat, 10°. F. increase in superheat 
decreases steam consumption 1 per cent. 

From 100°-200° superheat, 12° F. increase in superheat 
decreases steam consumption 1 per cent. 

From 200°-300° superheat, 14° F. increase in superheat 
decreases steam consumption 1 per cent. 

For each 1 per cent, moisture the steam consumption will be 
increased 2 per cent. 

For an increase in vacuum from 26 in. to 27 in. the steam con- 
sumption will decrease 5 per cent. 

For an increase in vacuum from 27 in. to 28 in. the steam con- 
sumption will decrease 6 per cent. 

For an increase in vacuum from 28 in. to 28J^ in. the steam 
consumption will decrease 3.87 per cent. 

For an increase in vacuum from 28J£ in. to 29 in. the steam 
consumption will decrease 5.75 per cent. 

For low-pressure turbines the following corrections may be 
made: 

Increase in vacuum from 26 in. to 27 in. decreases steam con- 
sumption 12 per cent. 

Increase in vacuum from 27 in. to 28 in. decreases steam con- 
sumption 13.75 per cent. 

Increase in vacuum from 28 in. to 28J£ in. decreases steam 
consumption 8.5 per cent. 

Increase in vacuum from 28J^ in. to 29 in. decreases steam 
consumption 11.25 per cent. 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 329 

For pressure, the increase of pressure of 10 per cent, between 
100 and 140 lbs. gauge pressure decreases steam consumption 
2 per cent. 

The increase of pressure of 10 per cent, between 140 and 180 
lbs. gauge pressure decreases steam consumption 1.95 per cent. 

The increase of pressure of 10 per cent, between 180 and 200 
lbs. gauge pressure decreases steam consumption 1.90 per cent. 

For low-pressure turbines 10 per cent, increase in pressure 
decreases the steam consumption 4 per cent. 

In any case if the values of the theoretical thermal efficiency 
for two sets of conditions are computed the ratio of steam con- 
sumption may be assumed equal to the ratio of the efficiencies 
and the equivalent steam consumption thus found. 

COMBINED ENGINE AND TURBINE 

Since the turbine is of especial value for low pressures and 
high pressures lead to certain troubles, steam engines have been 
used to handle the steam first, after which the steam is exhausted 
into turbines for use at low pressures. The turbines using 
exhaust steam are known as low-pressure turbines. In these 
the exhaust steam, at atmospheric pressure from engines or 
other apparatus, is utilized by the turbine. This combination of 
engine and turbine has resulted in very low steam consumptions. 

If the exhaust is not sufficient at times to drive the turbine, 
a set of blades is placed in the turbine on which high-pressure 
steam may be used in a high-pressure part in addition to the 
low-pressure steam. This is known as a mixed flow turbine. 

TOPICS 
Topic 1. — Derive the formula 

«>* = \2gJld, - i 2 )(l -y) + ^p 

What is y? What is the meaning of %<*. in the above expression and what 
would it stand for if the bracket (1— y) were omitted? How is y found? 

Topic 2. — Explain how the area F x is found for different points of a nozzle 
with a uniform pressure drop. Explain why a throat exists. What is the 
critical pressure? Explain the terms: throat, mouth, over expansion, 
under expansion. What is the effect of the last two? 

Topic 3. — Explain how the pressure in a tube at a given section may be 
found by experiment and how it may be calculated. Explain how this may 
be used to compute y. Sketch the form of apparatus used by Stodola and 
sketch the form of his curves. 



330 HEAT ENGINEERING 

Topic 4. — When are orifices or converging nozzles used? Sketch Rateau's 
forms. What fixes the length of a nozzle from entrance to throat and from 
throat to mouth? Give the steps taken in the design of a nozzle. 

Topic 5. — Sketch and explain the action of an injector. 

Topic 6. — Give the f ormulse for the following velocities : (a) at the throat 
of a steam nozzle of an injector; (6) at the mouth of a steam nozzle of an in- 
jector; (c) at a point within the combining tube for the steam; (d) at a 
point within the combining tube for the water; (e) at the throat of the 
delivery tube; (/) after impact in the combining tube for the mixture. 

Topic 7. — Write the two equations on which the action of the injector is 
based. What do these equations determine? 

Topic 8. — Derive the formula for the delivery tube : 



4 / z_ 

\ 20d t 



Topic 9. — Explain the action of a jet on a blade and prove the formulaB 

_ mw 
9 

P = —{w a cos a — Wb). 

Topic 10. — Explain the action of a jet on a curved blade. Sketch the 
inflow and outflow triangles and prove that 

work per pound = - (w a cos a — w' a cos ct')wb 

What is meant by velocity of whirl? What are impulse and reaction 
turbines ? 

Topic 11. — Sketch the inflow and outflow triangles and give all resulting 
trigonometric relations. 

Topic 12. — Deduce the condition for maximum efficiency when a is fixed 
and = 180 - £'. 

Topic 13. — Deduce the condition for maximum efficiency with a fixed value 
of |8 and p'. 

Topic 14. — Explain the meaning of pressure compounding and velocity 
compounding. Explain the construction of a one-stage and two-stage ve- 
locity diagram with friction and without friction. Explain how to find the 
work and kinetic efficiency from this. 

Topic 15. — Explain by diagrams the peculiar features of the turbines given 
in text. Sketch the curves showing the variation of pressure and velocity 
through the turbine. 

Topic 16. — Give the expressions for the various component efficiencies of a 
turbine and the expression for the overall efficiency. Explain these and the 
method of computing them. Explain the similarity of thermal action of an 
engine and steam turbine. 

Topic 17. — Sketch a Mollier chart and on it show why a reheat factor 
exists and derive the expressions 



STEAM NOZZLES, INJECTORS, STEAM TURBINES 331 

tan 5 = kT = - 
s 

U 



i 

R.H. = - 



<r 



1-^ 
Ti 

Topic 18. — Outline the method of design of a turbine to give a definite 

power. 

PROBLEMS 

Problem 1. — Find the velocity of discharge at throat and mouth of a 
nozzle operating with dry steam between 155 lbs. gauge pressure and 70 lbs. 
gauge pressure. Find the area of the nozzle to care for 30 lbs. of steam per 
minute. 

Problem 2. — Draw a curve representing the change of area from 2 sq. in. 
to 0.25 sq. in. in 0.5 in. and then an enlargement to 2 sq. in. in 3 in. Find 
the pressure along this nozzle if the initial pressure is 135 lbs. absolute and 
the steam is superheated 230° F. 

Problem 3. — Find the size and shape of a Rateau nozzle to deliver 2500 
lbs. of steam per hour from a pressure of 25 lbs. gauge to a pressure of 15 lbs. 
gauge, xi = 0.99. 

Problem 4. — Design an injector to feed 10,000 lbs. of water per hour into a 
boiler under a gauge pressure of 220 lbs. The lift is 5 ft. Make sketches of 
nozzle, combining tube and delivery tube. 

Problem 5. — An injector operating with steam at 175 lbs. gauge pressure 
is to force water into a boiler in which the gauge pressure is 275 lbs. How 
much water is lifted per pound of steam ? 

Problem 6. — Prove that it is possible to feed a boiler under 120 lbs. gauge 
pressure by means of steam at 25 lbs. absolute pressure. 

Problem 7. — Find the kinetic efficiency of a single-stage, impulse turbine 
with cos a -1 = 0.93 without friction and with friction. Find the efficiency if 
= 45° and 0' = 135°. 

w a — 2700 ft. per sec. 

Problem 8. — Find the kinetic efficiency of a two-stage impulse turbine with 
cos oT 1 =0.93. Use no friction for first assumption. w a = 2700 ft. per 
sec. Assume correct values of / to suit velocities and find the efficiency 
with friction. 

Problem 9. — Solve Problem 8 assuming that a. = on. 

Problem 10. — Find the reheat factor for four stages on a Curtis turbine 
when rjk = 0.75, r) n = 0.90. The pressure range is from 175 lbs. gauge 
pressure, 100° F. superheat to 29 in. vacuum. 

Problem 11. Find the pressure on the four stages of Problem 10 if the re- 
heat factor is 1.03. What is the thermal efficiency? What are the various 
efficiencies? What is the probable steam consumption? Find leading 
dimensions. 

Problem 12. — Find the reheat factor for a twenty-stage Rateau turbine 
with the same values of efficiency and range of pressure as that given in 
Problem 10. 



332 HEAT ENGINEERING 

Problem 13. — Determine the leading dimensions of a 200-kw. De- 
Laval turbine generator to operate between 150 lbs. gauge pressure and 
2 lbs. gauge pressure. Give the probable steam consumption. 

Problem 14. — Determine the leading dimensions, number of stages, reheat 
factor and probable steam consumption of a Parsons turbine to develop 3000 
kw. with pressure of 175 lbs. gauge and 50° superheat to a 29-in. vacuum. 



CHAPTER VIII 
CONDENSERS, COOLING TOWERS AND EVAPORATORS 

The condensers used in practice are of different forms. They 
are of the surface form when it is desired to use the condensate 
and the cooling water is improper for boiler feed. When the 
water used for cooling is satisfactory the jet type may be used. 

Fig. 161 shows the usual form of surface condenser. In this 
steam enters at A and passes over the tubes, which are filled by 
water which enters at B and returns to C where it is discharged. 




Fig. 161. — Surface condenser. 



The baffle plate D is used to spread the steam over the surface 
of the tubes and prevent short circuiting. The condensed steam 
is drawn off at E and since the air entrained in the boiler feed 
would destroy the vacuum as it collects, means must be taken 
to remove this air. The pump which takes the condensate from 
this space of low pressure to the atmosphere is made sufficiently 
large to handle this air as well as the water and is therefore called 
an air pump. 

The air present with the steam is a very small percentage of 
the mixture as it leaves the engine or turbine, but as the steam 
condenses the non-condensable air becomes a larger part of the 
remaining mixture of vapor and air. Air being heavier than 

333 



334 HEAT ENGINEERING 

steam, the mixture containing the air naturally travels to the 
lower part of the condenser where if it is not removed fast enough 
it will so fill the space that steam cannot reach the condensing 
surface at this place and so the efficiency of the condenser is 
diminished. 

Modern practice has demanded such low pressures in the con- 
denser that a separate pump is necessary to remove the air alone 
from the bottom of the condenser at F while a small pump is used 
for the condensed steam at E. The air pump is then called a 
dry air pump, and by taking this air through a space cooled by 
tubes containing the coolest water available the volume of this 
air is decreased by the increase of air pressure due to the decrease 
of the vapor pressure. 

PRESSURES IN A CONDENSER 

If p c is the actual pressure in the condenser shown by the 
vacuum gauge, this pressure is equal to the sum of the pressure 
due to the steam and that due to the air; Since the steam is in 
the presence of water, the steam pressure p s is the saturation 
pressure corresponding to the temperature and the air pressure 
p a is the difference of these. 

Pa = Pc - Ps (1) 

The pressures p a and p s are known as partial pressures. 

If now by passing this mixture of vapor and air through a cold 
space p s is made smaller, the pressure p a will be made so much 
greater that the volume of the air will be materially decreased. 
This requires a smaller air pump. 

The pressure present in a condenser with no air corresponds 
to the saturation temperature, but as air is added this pressure 
rises, or for a given pressure the temperature of the mixture will 
be lowered as more and more air is permitted to enter. 

The amount of air present has been the subject of much in- 
vestigation. G. A. Orrok in the Transactions of the American 
Society of Mechanical Engineers, Vol. xxxiv, p. 713, etc., has 
shown that although Croton water contains 4.3 per cent, of air by 
volume, this is decreased to about 0.93 per cent, when heated in 
open heaters to 187° F.; 0.0151 per cent, of the 0.93 per cent, is 
air mechanically mixed and 0.916 per cent, is in solution. This 
air is referred to its volume at atmospheric pressure. Although 
the feed water contains 0.93 per cent, of air the water in the 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 335 



hot well contains 0.269 per cent, of air, showing that for eaoh 
cubic foot of feed water 0.00661 cu. ft. of air at atmospheric 
pressure had been liberated. At the low partial air pressure 
present in the modern condenser this is increased a hundredfold 
in volume. Mr. G. J. Foran in the discussion of this paper gives a 
curve from Castell-Evans Physico-Chemical tables showing the 
percentage absorption at different temperatures when the total 
pressure outside is 760 mm. This is given in Fig. 162. Although 
this does not agree with the values found by Orrok it shows how 
the solubility changes with the temperature. 



S 0-022 


































© 

& 0.018 

£ 0- 016 

ft 

5 0.014 

2 0.012 
o 

H 0.010 
© 

» 0.008 

a 

o 


































































































































































































































o 

° 0.004 

o 

1 0.002 

.a 



































































































212 U F 



192 



172 



152 132 112 92 

Temperature in Degrees F 



72 



52 



Fig. 162. — Amount of air at atmospheric pressure absorbed by 1 cu. ft. of 
water at various temperatures given by Foran from Winkler's data. 

Orrok found that with turbine units of 5000 to 20,000 kw. 
the amount of air at atmospheric pressure and temperature 
varied from 1 cu. ft. per minute when the units were tight to 15 
or 20 cu. ft. with ordinary leakage, and 40 to 50 cu. ft. per minute 
when the units were not tight. This shows that such air is due 
to leakage and not to the air contained in the feed water. The 
minute holes in the shells, heads and expansion joints are difficult 
to detect, but the saving from the elimination of these is much 
greater than the cost of that elimination. Although this leakage 
may amount to a considerable quantity, the shutting down of the 
air pump does not mean an immediate loss of vacuum. In a 
test mentioned by Mr. Foran a condenser caring for 9000 kw. lost 
only 0.3 in. in 30 minutes. 



336 



HEAT ENGINEERING 



Orrok found that there was too 
much drop of pressure in the con- 
denser proper. To reduce this 
the condensers should be made 
broad and shallow. The drop 
amounted to about 0.2 in. of 
mercury, increasing with the load 
to 0.6 in. He found that the 
power required for the air pump 
condensers amounted to about 25 
i.h.p. on the steam end for loads 
between 6000 and 10,000 kw. 
The mechanical efficiency of the 
dry air pumps was about 50 per 
cent, and the volumetric efficiency 
working between 34 lb- and 15 
lbs. was very low indeed, due partially to the low value of the 
clearance factor and to the large ratio of compression. 




Fig. 163. — Shell and partitions of 
a Wheeler dry tube condenser. 




Y Condensate 

Uniflux Contraflo 

Fig. 164. — Sections of uniflux and contraflo condensers in which steam pas- 
sages grow small as steam condenses. 

During these tests the absolute pressure was about 1J^ in. of 
mercury and was obtained with 15° F. to 20° F. rise in the cooling 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 337 



To Dry Air Pump 



Steam 



water at 35° F. The condensed steam was 10° above the tem- 
perature of the outlet water although the steam space might be 
30° above this. 

To better the cooling effect of the surface of condensers the 
velocities of the water and steam across the surface should be 
made high. This was explained in Chapter III. 

The attempt has been made to keep the tubes partially dry 
from the drip of the condensate in the Wheeler Dry Tube Con- 
densers, Fig. 163, by putting partitions across the condenser. 
The condenser tubes have been omitted from the figure so that 
the partitions may be shown 
clearer. This of course in- 
creases the velocity of the 
steam and so improves the 
heat transmission. This is a 
development of the English 
contraflo condenser. In 
Weir's Uniflux Condenser, Fig. 
164, the shell is so made that 
the velocity of steam remains 
nearly constant as it is con- 
densed and crosses the tubes. 
As the volume decreases, due to 
the condensation of part of the 
steam, the space through 
which the vapor passes gradu- 
ally diminishes. The uniflux 
condenser shown at the left of 
Fig. 164 is almost an equivalent 
of the contraflow in a circular 
shell, the partitions of the 
contraflow guiding the steam FlG - 165.— Barometric jet condenser. 
into a channel of decreasing area. 

When the condensing water is suitable for boiler feed or where 
it is not necessary to save the condensate, jet condensers are often 
used because they are cheap and may be effective. In them the 
water and steam are brought into intimate contact. Fig. 165 
shows the type of jet condenser known as the barometric counter- 
current condenser 

In this steam enters at A and meets the numerous streams of 
condensing water by which it is condensed, and falls to the tail 

22 




Barometric Tube 
34 ft. Long 



338 HEAT ENGINEERING 

pipe B and finally discharges into the hot well. The cold water 
enters at C and discharges through numerous orifices into the 
highest trough D from which it discharges into the other troughs 
in succession. Air is removed by a dry air pump connected at 
E and all air has to pass through streams of the coolest water 
before leaving the condenser head. In this way p s is made as 
small as possible, and p a as great as possible. 

CONDENSER DESIGN 

The first point to be settled in the design of a condenser is the 
temperature of the cooling water and the vacuum desired. In 
most cases there will be a rise of about 20° in the cooling water, 
its outlet temperature will be 10° to 30° less than the tem- 
perature in the condensing space and the temperature of the hot 
well will be less than the temperature in the condenser by about 
10° or 20°. A standard assumed by condenser designers for 
high vacuum work is a 15° rise in the cooling water and a tem- 
perature of outlet of 15° below the temperature corresponding 
to the vacuum carried. The greater the difference in temperature 
between the water and the steam, the more effective the surface 
will be. If then U, the inlet temperature, is known, t 03 the tem- 
perature at outlet, is assumed and finally t s of the steam is as- 
sumed. From this p s is known, and then if the amount of air 
present is known p a may be computed and from it p c , the con- 

denser pressure expected on the condenser. The ratios of — 

Pc 

will be about 0.8 or 0.9. Knowing this, the amount of water 
per pound of steam is given by 

oJ f7^ (2) 

where G = pounds of cooling water per pound steam 
i = heat content of steam entering condenser 
q\ = heat of liquid at temperature of hot well 
q' = heat of liquid at temperature of outlet 
q'i = heat of liquid at temperature of inlet. 

If this is a jet condenser 

<A = q'o (3). 

These may be taken within 5° of the temperature in the head. 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 339 

After G is found the next calculation is that of the surface 
required for a given amount of steam. In practice \y± to 2 
sq. ft. of surface are used per kilowatt. 

If W is the weight of steam per kilowatt-hour output the total 
weight of water used will be 

total water per hr. = G X W X kw. (4). 

The quantity of heat is 

Q = G X W X kw. X fe'o - q'i) (5) 

Using the formulas of Chapter III for the surface the following 
results : 

K mean AT heat per sq. ft. ^ ' 

In computing the heat per square foot the velocity of the water 
may be taken as 4 ft. per second. 

The tubes are arranged in nests so that the velocity of the water 
is that desired and then the spaces between the tubes are made 
to give a uniform velocity by means of the partitions. 

After this the power and size of the water pump may be com- 
puted by assuming a pressure of say 5 lbs. for the resistance in 
the condenser tubes. The power and size of the air pumps are 
then found. 

These various points will be brought out in the problems below. 

PROBLEMS 

Problem 1. — Suppose that 20 cu. ft. of air per minute are found to be 
present in the exhaust of a 10,000-kw. turbine. At a temperature of 75° 
in the condenser chamber and a pressure of 1 in. what would be the values of 
p a and p a ; and what would be the amount of air to be cared for by an air 
pump if the air is cooled to 60° F. before entering the cylinder? Steam per 
kilowatt hour = 14 lbs. 

Steam pressure at 75° = p s = 0.4289 lb. per sq. in. 

Total pressure = p c = 0.49 lb. per sq. in. 

Partial air pressure = p c — Ps = 0.0611 lb. per sq. in. 

Steam pressure at 60° = 0.2561 lb. per sq. in. 

Air pressure = 0.2339 lb. per sq. in. 

It will be seen that by the cooling of this air the pressure has been increased 

fourfold. 

15 
Volume of air per min. = 20 X Q 2 ooq = 1283 cu. ft. 



340 HEAT ENGINEERING 

The pressure of the air in the dry air pump is raised from y± lb. to 15 lbs. 
and the clearance factor and leakage factor would be computed as in the 
case of the air compressors of Chapter IV. 

1283 
^ clearance factor X leakage factor ^ ' 

Jh _n piFi I 1 " W~ n / 

in -P- n - 1 leakage factor 33000 ( 8 ) 

Problem 2. — Find the amount of cooling water for the condenser of 
Problem I if the inlet water is at 50° F. 

k = 50° 

U = 75° 

t assumed 65° 

t h = 55° 

Value of 2 75 = 43.1 + 1049.2 X 0.895 = 983.1 
(0.895 was assumed from turbine problem in Chap. VII) 
„ 983.1-23.1 „,„ 
G= 33.1- 18.1 =641bs ' 

Total weight water = 10,000 X 14 X 64 = 8,950,000 lbs. per hr. 
Total volume per minute = 2400 cu. ft. or 17,800 gal. per min. 

Problem 3. — Find the surface required for the condenser above using 
admiralty tubing and assuming clean tubes with a water velocity of 4 ft. 
per second. 

Ah = 75 - 50 = 25 

At 2 = 75 - 65 = 10 
H i/ 8 (25-10) "1 8/7 /1.875\9* 
Mean M = L(26)* - (loH = lol62; = 16 ' 4 

630 X 1.0 X (^|) 2 0.98V4 

K =~ (16.4)H -= 67 ° 

Total heat per hr. = 8,950,000 X 15 = 670 X F X 16.4 

F = 12,200 sq. ft. 

Increasing this 20 per cent, gives 

F = 14,600 sq. ft. or 1.46 sq. ft. per kw. 

In the N. Y. Edison Plants 1.1 sq. ft. are used per kilowatt in one of 
their new turbines. At the Interborough Station in New York 1.67 sq. ft. 
are used, and at the Fisk Street Station in Chicago 2.08 sq. ft. are used. 

The cost of operating the auxiliaries of condensers amounts to 
33^ per cent, of the power of the engine or turbine, about 3^2 P er 
cent, being taken for the air pump and 3 per cent, for the cir- 
culating pump. 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 341 

In the case of turbines a low vacuum can be used because the 
toe of the diagram is used, the expansion of the steam being of 
value to the lowest point. On account of the steam engine be- 
ing unable to use the toe of the expansion diagram, due to the 
enormous volume required, the gain in power for part of the 
volume by decreasing the pressure is not enough to pay for the 
greater expense in operating the pump and the reduction in the 
temperature. This was explained in Chapter II. 

COOLING TOWERS 

Where cooling water is not obtainable as 'in the center of a 
large city and it is desired to operate the apparatus condensing, 




Fig. 166. — Cooling tower. 

cooling towers are used. These are of different forms. In all 
of them air is allowed to come in contact with the warm condens- 
ing water which has been divided by some device into fine streams 
or sheets. In some cases the water is allowed to trickle over 
mats or screens of wire or is discharged over vitrified tiles by a 
distributing head. In other cases the water is discharged over 
boards. The general plan is to get the water into thin sheets or 
drops so as to expose a large area to the action of the air. To 
bring the air in contact with the water it is forced through by a 
fan or by a flue or chimney built on top of the tower to produce 
a draft by the warm air which sucks fresh air into the bottom of 



342 HEAT ENGINEERING 

the tower. This air becomes heated from the warm water and 
has its moisture content so increased that the evaporation of the 
water to satisfy this, added to the heat required to warm the air, 
so cools the water that it reaches the bottom of the tower at a 
temperature suitable for the condenser. 

Fig. 166 illustrates one form of tower. The fans A are driven 
by an engine or motor and force the air through the wire 
screens B, called mats. The water enters the troughs C-C from 
the pipe D and is distributed into tubes E which have their upper 
part removed, forming troughs which overflow and distribute 
their water over the mats. By continuing the casing 30 or 40 ft. 
above the troughs there would be sufficient chimney effect to 
draw considerable air, although for maximum capacity fans are 
required. Cooling towers take about 5 per cent, of the power of 
the apparatus being operated by the condenser, 2 per cent, for 
the fan and 3 per cent, for circulating the water. 

The amount of moisture that will be taken up by air depends 
on the amount of moisture present in the air. All liquids dis- 
charge particles into any space around them forming a vapor of 
the substance in the space above the liquid. Particles also fall 
back from the space into the liquid. The number of particles 
leaving the liquid is greater than the number falling back until 
the pressure exerted by this vapor is equal to the vapor tension 
or saturated steam pressure in the case of water, corresponding 
to the temperature of the liquid. At this time the space above 
is said to be saturated and the weight of vapor per cubic foot is 
the weight of a cubic foot of saturated vapor given in the tables. 
If the amount of moisture in the air is not equal to this, the air 
is not saturated with moisture and the ratio of the weight of 
moisture per cubic foot to the weight when saturated is called 
the relative humidity. 

P = ^ (9) 

m s 

It happens that the ratio of the pressure exerted by this 
moisture compared with that at saturation is practically the same, 
since the pressure of a perfect gas is proportional to the mass 
when the volume and temperature are constant. Of course 
this is not a perfect gas and the statement is not absolutely true. 

p = v i (10) 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 343 

To find the relative humidity a hygrometer is used. The 
common form used in engineering work is that consisting of two 
thermometers, one of which has its bulb surrounded by a damp 
piece of wicking. As this is twirled at a high rate in the air the 
evaporation of the water on this bulb lowers its temperature, the 
amount of lowering being dependent on the relative humidity. 
The fall of temperature is then compared with the temperature 
of the dry bulb and the barometric reading; and from tables 
or formulae the relative humidity is found. 

Carrier proposes the formula 

p' Bar. - y t - V 

9 Vt Vt 2755 - 1.28*' K J 

p' = pressure corresponding to wet bulb temperature 
p t = pressure corresponding to dry bulb temperature 
Bar. = barometric pressure 
t = temperature of dry bulb 
t' = temperature of wet bulb. 

If now the air enters the cooling tower at a temperature ti 
and of relative humidity pi the amount of moisture per cubic 
foot is 

raipi 

where ra x = weight of 1 cu. ft. of steam at temperature t\. 

One cubic foot of entering air is considered in problems of 
the cooling tower as it leads to simple calculations. 

If the temperature of the warm water is U, the temperature 
t 2 of the air leaving may be taken as 10 to 20° below U but satu- 
rated on account of the intimate mixture. The water leaving 
the tower will probably be-above the temperature ti of the en- 
tering air although it might be equal to it in some cases. Assume 
h to be 10° F. or 20° F. below the temperature t of the outlet 
water. With intimate mixture h may correspond to t and t 2 
to t { . 

The moisture per cubic foot leaving the tower is ra 2 but the 
number of cubic feet have been changed on account of the. 
changes of temperature and pressure. Let the weight of air be M . 
M = (bar. - P ,p l )lAAV l = (bar. - p 2 )144F 2 ( 

BT\ BTi 

Bar. = barometric pressure in lbs. per sq. in. 
Pi = relative humidity at 1 



344 HEAT ENGINEERING 

Pi = saturation pressure at 1 
_ i = absolute temperature at 1 
Vi = volume at 1 = 1 cu. ft. 

The volume of air at outlet corresponding to 1 cu. ft. at 
entrance is 

bar. — p 2 Ti 

The weight of moisture evaporated per cubic foot of original 
air is 

m e = m 2 V 2 — pimi (14) 

If M a = weight of cooling water cared for by 1 cu. ft. of air 
at entrance, the weight of water remaining is 

M a — m e 

The late Prof. H. W. Spangler solved this problem by equating 
energies at entrance and exit and hence it is necessary to find 
the energy brought in by each substance above 32° F. 

I. Energy in water at bottom of tower above 32° F. = 

{M a - m e ) q' (15) 

II. Energy in air at entrance above 32° F. = 

A (bar. - pipi)144 X 1 T7 n „. 

III. Energy in moisture entering above 32° F. = 

pim-i [ii — Apiv'"] = pmiii — Api X 1 (17) 

i is heat content of superheated steam at pressure pipi and of 
superheat of h — t s degrees. 

IV. Energy in water at top of tower above 32° F. = M a q'i (18) 

V. Energy in air at exit above 32° F. = 

A (bar. - p 2 )144F 2 , . 
1.4 - 1 u * 2 U ; 

VI. Energy in moisture leaving above 32° F. = 

m 2 F 2 W2 + r 2 ] — Ap 2 V 2 
or better U = m 2 7 2 W2 + p 2 ]. ( 2 0) 

VII. Work done by air in changing 1 cu. ft. to V 2 cu. ft. at 
barometric pressure = 

A bar. X 144 X (V 2 - 1) (21) 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 345 

Now the sum of the energies entering must equal the sum 
of those leaving plus the work: 

II + III + IV = I + V + VI + VII (22) 

In this equation the only unknown is M a and this may be 
found. From M a the number of cubic feet of air required for 
a given installation may be known and from this the size of the 
fan, power required, size of tower and other data may be 
ascertained. 

This is the best method for solving such a problem as other 
methods of attack are open to objections since the temperature 
at which events take place is not known. 

PROBLEM 

Suppose air at 70° gives a wet bulb temperature of 58.5° F. and is used in 
a cooling tower with hot water at 140° F. The barometer is 14.7 lbs. The 
air is so mixed that it leaves at 140° F. while the leaving water is cooled to 
80° F. How much air would be required to cool 30,000 lbs. of water per 
hour. 

P58-5 = 0.2428 
p 70 = 0.3627 

70 - 58.5 
2755 - 1.28 X 58.5 

0.5 





0.2428 14.7 - 0.2428 


PI 


~ 0.3627 ~ 0.3627 




= 0.668 - 0.171 = 0.497 = 


m x 


= 0.001152 


P\m x 


= 0.000576 


mi 


= 0.00814 


V2 


= 2.885 


v 2 


14.7 - 0.5 X 0.3627 v 601 


14.7 - 2.885 X 531 



1.39 
>± 

m 2 V 2 = 0.00814 X 1.39 = 0.0113 
m e = 0.0113 - 0.000576 = 0.0107 

Energy in water at bottom = (M a - 0.0107)48.1 = 48.1M - 0.515. 

u. • • 1 (14.7-0.1814)144 TT a „ n TT 

Energy in air at entrance = -==z ^-j Ua = 6.70 — U 3 2- 

Energy in moisture = 0.000576 [19.1 + 1061.8 + 19 X 0.43] - | — X 

L778 

0.1814 X 144 1 = 0.600. 

(Moisture at entrance is under a pressure of 0.1814 lb. and at a temperature 
of 70°. 0.1814 lb. means a saturation temperature of 51° or the moisture is 
19° superheated. From the curves of specific heats in Chapter I, c p is 0.43.) 



346 HEAT ENGINEERING 

Energy in water at top of tower = M a X 108.0. 

j? .... 1 [14.7 - 2.885J144 X 1.4 TJ __ u 

Energy in air at top = ==~ tt-j U32 = 7.65 — C/32. 

Energy in moisture at top = 0.0113[108.0 + 947.5] = 11.9. 

Work done by changing 1 cu. ft. of volume to 1.4 cu. ft. of volume at 

atmospheric pressure = ==^ X 14.7 X 144(1.4 — 1) = 1.09. 

6.70 - C/32 + 0.600 + M a X 108.0 = 48.1Ma - 0.515 + 7.65 - U S2 

+ 11.9 + 1.09 

1 2 82 

M ° ' M¥ - a214 

This is the amount of water cared for by 1 cu. ft. of air at entrance, the 
reciprocal, 4.7 cu. ft., being the number of cubic feet of air required to care 
for 1 lb. of hot water. The amount of water evaporated per cubic foot 
of air has been found to be 0.0107 lb. This quantity is 5.2 per cent, of 
the water supplied (0.214 lb.) per cubic foot of air. 

In the problem 30,000 lbs. of water are passed through the 
cooling tower. This requires 4.7 X 30,000 or 141,000 cu. ft. of 
air per hour. The evaporation will amount to 5.2 per cent, of 
the weight of water or 1.600 lbs. of water per hour. Had this 
cooling tower been supplied with water at a lower temperature 
the evaporation would .have been much less but the amount of 
air would be greater. With the temperature of the condensing 
water from a condenser at 140° F. (this temperature is much 
higher than that found in practice) with a supply temperature 
of 80° F., 16 lbs. of water would be required per pound of 
steam so that 30,000 lbs. of water would be used with 1880 lbs. 
of steam. If this condensed steam were used in a cooling tower 
it would more than care for the loss by evaporation and if used 
for boiler feed, the make-up water to be bought would be less than 
the feed water saved. In all cases where the condensate can be 
used either for boiler feed or condensing water the cooling tower 
reduces the bill for water. If the water cannot be used it will 
be found in most cases that the sum of the make-up water and feed 
water will be less than the steam necessary with a non-condensing 
plant. In all cases the heating of the air and the doing of ex- 
ternal work makes the amount of evaporation in the tower less 
than the amount of condensation cared for by the cooling 
apparatus. 

It will be well to keep in mind that although the air entering 
the lower part of the cooling tower be saturated with moisture, 
due to rain or snow, the greater moisture content for the warm 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 347 

air makes evaporation possible even in this case. The warming 
of the air also removes heat and cools the water. 



SIZE OF TOWER AND MATS 

The size of the tower necessary to be used may be found by 
assuming the air to pass through at a velocity of 700 ft. per min- 
ute and the area of the mats or cooling surface may be found by 
allowing 200 B.t.u. per hour per square foot of surface with 10° 
cooling to 700 B.t.u. per hour per square foot with 35° cooling. 
This may sometimes be expressed in terms of the water, 1 sq. 
ft. for 25 lbs. of water per hour. This is independent of the 
amount of temperature change. 

For the problem above the net area of the horizontal cross section of the 
tower would be 

F - - ifs - 3 - 36 sq - ft - 

The area of the mats would be 

_, 30000(140 - 80) 1Qnn 

m = — looo* = sq# 

* This is assumed for the large drop of 60° F. 
F m - *%?■ ■ - 1200 sq. ft. 

In planning this tower care should be taken to prevent the 
water from falling free. It must be kept in contact with the 
mats. There must be care in arranging the passages to prevent 
air from taking any path which does not bring it into contact 
with the water. 

Towers cost from $1.00 to $6.00 per kilowatt capacity or as 
much in some cases as the condenser equipment. 

The power to drive the fan is about 2 to 5 per cent, of the 
engine power. In the case above the practical use of the fan is 
to give velocity only, as the drop in pressure is practically nothing. 
For this reason the formula for the power of the fan need only 
consider the kinetic energy of the air, the friction head being 
considered as equal to the velocity. This gives 

141,000 X 14.7 X 144 /700\ 2 1 
work per minute = 2 X 60 x 5 3.35 X 531 Uo / UA 

= 750 ft. -lbs. per min. 



348 HEAT ENGINEERING 

SPRAY NOZZLES AND PONDS 

Another device often used to cool water is the spray foun- 
tain. Nozzles are supplied with hot water from a main and the 
velocity of discharge is made to divide it into a fine spray. The 
water is then caught in a reservoir from which it is taken to the 
condenser. With these nozzles the water may be cooled 15° 
when the atmosphere is 25° to 30° below the temperature of the 
hot water while 20° may be obtained with a 40° difference. These 
nozzles are placed about 8 ft. apart and are usually fitted to 3-in. 
pipes in which the velocity should be about 5 ft. per second. 
In this way each nozzle will care for about 60,000 lbs. of water 
per hour. 

In some plants large cooling ponds are used to cool the 
water by surface evaporation. With these the hot water is 
discharged at one end of the pond and the cooling water is taken 
off at the other. Thomas Box, many years ago, recommended 
that 210 sq. ft. of cooling surface be used per nominal horse- 
power if the engine was operated for 24 hr. per day. This 
was with engines using more steam than those of to-day and 
this might be reduced to say 120 sq. ft. per 24 h.p.-hr. 
per day, or about 5 sq. ft. per horse-power hour during the day 
of 24 hr. W. B. Ruggles, in the Journal of the A.S.M.E. 
for April, 1912, gave a test of a cooling pond of 288,000 sq. ft. of 
a depth of 5.38 ft. in which he found a transfer of 3.67 B.t.u. 
per square foot of water surface per hour per degree difference in 
temperature between air and water, and that 120 sq. ft. of surface 
per horse-power was sufficient when 16 lbs. of steam were used 
per horse-power hour. 

ACCUMULATORS AND EVAPORATORS 

Regenerator accumulators are devices used for the retention 
of surplus heat until needed when an intermittent supply is 
given off by a machine and it is desired to use this in another piece 
of apparatus. They were planned by Rateau to be used with 
low-pressure turbines which were operated by the exhaust 
steam from engines, the operation of which was intermittent as 
is the case with rolling mill engines or from steam hammers. 
One of his forms is shown in Fig. 167. It consists of a vessel A 
made of steel in which there is a horizontal partition at the center. 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 349 

A number of elliptical flues B.B. are placed in the tank. These 
receive steam from the pipe D and manifold C and steam is dis- 
charged through %-in. holes in the flue walls into the water carried 
in the chambers. This water may enter the flues. The steam 
heats the water which is introduced from the float box E as soon 
as the water level is lowered. The steam entering from D may 
also enter the top of the accumulator by the check G and pass 
out to the turbine through F. If however the intermittent supply 
is reduced or stopped momentarily, the pressure in the accu- 
mulator falls and the warm water begins to evaporate, thus main- 
taining the steam supply. Of course if the supply of steam is 
discontinued for a long time, the turbine receives its steam 
through a reducing pressure valve from the boiler main. Baffle 
plates are used above the elliptical steam flues to cause the water 
and steam to mix and to dry out the steam leaving the accu- 




Fig. 167. — Rateau accumulator. 

mulator. In some Rateau Accumulators large masses of iron 
in the form of trays are used to absorb the heat and supply that 
necessary to evaporate the water when the supply is reduced. 

Evaporators are used for the purpose of concentrating liquors or 
solutions and for the production of distilled water or other liquid 
although when used for this latter purpose they are known as 
stills. Fig. 168 shows one form of evaporator. In this one a 
liquid to be evaporated is carried to the line A in a tank B con- 
taining a set of tubes C held between two tube plates. The space 
D between the tube plates is separated from the remainder of the 
shell. This space is supplied with steam or some other hot vapor. 
The vapor gives up its heat to the fluid within the tubes and is 
condensed, the condensate leaving at E. If the pressure in the 
chamber above the level of the liquid is such that the boiling 
temperature of the liquid is below the temperature of the vapor 
entering the space D from the pipe F, the liquid will boil and its 



350 



HEAT ENGINEERING 



vapor may be used to boil liquid in a second chamber in which 
the pressure is maintained still lower than that in the first cham- 
ber. This operation may be repeated as shown in the figure, 
the vapor from the last chamber B" passing to the condenser H 
in which the pressure is maintained at a low point by the air pump 
I. In many cases where these are used for the concentration of 
liquors as in sugar making, the dilute solution enters at J and as 
it becomes more concentrated it sinks to the bottom of B and is 
passed over to the upper level of K r of the next evaporator. 
Since the pressure is lower in C than in B this action will take place 
and be regulated by opening a valve in the line. This action is 
carried on throughout the system. 




Fig. 168. — Triple effect evaporator. 

Where three of these are used as shown in Fig. 168, the arrange- 
ment is known as a triple effect evaporator, a single one as a 
single effect, two as a double effect, four as a quadruple effect. 

Assuming the arrangement as shown, suppose M pounds of 
liquor enter at J. Of this ei pounds are evaporated and M — e\ 
pounds pass on to the second effect to be evaporated. Of this 
e 2 are evaporated and M — (ei + e 2 ) are passed on and e 3 are 
evaporated and M — (e x + e 2 + e 3 ) are drawn off as a con- 
centrated solution. 

In the first effect suppose 1 lb. of vapor is admitted at F. 
This is condensed and the condensate is removed at E. This 
weighs 1 lb. ei pounds of vapor are admitted to F' and d 
pounds of vapor are condensed here. e 2 pounds are introduced 
and condensed in the third stage. 

In working out the condensation and evaporation in the differ- 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 351 

ent effects the method of procedure is to start with the one effect 
and work to the others. In these various tanks the amount of 
heat loss must be allowed for and if the size of the tanks be 
assumed the amount of heat loss per hour may be computed for 
each from the value K for coverings as given in Chapter III. 
The pressures in the various tanks are assumed so as to give the 
proper temperature differences. These may be made 20° for 
each stage. 

The condensates from the various stills could be discharged 
through a feed heater and this heat could be added to the feed 
entering the first effect so that the condensates would leave at 
the temperature of the fresh liquid. The same could be done in 
theory with the condensate from the condenser. The operation 
of the triple effect depends on a difference in the temperatures of 
the various effects and hence in any problem the initial steam 
pressure pi and temperature h and the final temperature t c 
would be assumed and with them the various intermediate tem- 
peratures, from the temperatures of the condenser and the tem- 
perature of the condensate. For a triple effect, the temperature 
of the steam in the first effect is h in the supply and t 2 in the dis- 
charge while £ 3 is the temperature of the discharge from the 
second and t c is the temperature of the discharge from the last 
stage. The temperature of the cooling water and weak liquor 
is U and that of the warm circulating water is t while the con- 
densate in the condenser is h. The following conditions hold for 
the various stages if 10 per cent, of the heat entering is assumed 
to be lost in radiation. 

First Stage — Heat Entering: 

With 1 lb. steam Q = q\ + r 1 (23) 

With ei + e 2 + e 3 lbs. feed Q = (ei + e 2 + e^)q f i 

+ lfa'i - q\) + eite'2 - q'i) + c 2 (q' 3 - q\) + 

e 3 (q\ - q'i). (24) 

(Feed heaters are used to reduce condensates to U) 
Heat Leaving: 

With e± lbs. evaporation Q = ei(q' 2 + r 2 ) (25) 

With 6 2 + e 3 lbs. feed to next evaporator Q = (e 2 + e 3 )q' 2 (26) 

With radiation Q = Kofe'i + rj (27) 

With 1 lb. condensate Q = q\ (28) 



352 HEAT ENGINEERING 

The equation for heat balance is 

q'i + fi + q\ — q'i + e x q' 2 + e 2 g' 3 + e 3 q\ = q\ + Ho (q'i + r i) 
+ 61(^2 + r 2 ) + (e 2 + e 3 )q' 2 

0.9(^1 + n) - </; = €l r 2 + e 2 (?'2 - ?' 3 ) + e 3 (q' 2 - q' h ) (29) 

The equation for the second stage in the same manner is 

= 6![0.9r 2 - 0.1^2] - e 2 [q' B + r 3 - q f 2 ] - e 3 [q'z - q' 2 ] (30) 
The third stage gives 

= e 2 [0.9r 3 - 0.1 g' 3 ] - e s [q' c + r c - q' 3 ] - m q f c (31) 

These three equations are sufficient to find the quantities 
e h e 2 and e 3 . 

In the last effect the final withdrawal of the concentrated 
liquor will give the subtractive term in the last equation: 

m q' c (32) 

m is very small compared with the other terms. 

PROBLEM 

As a problem, suppose water at 65° F. is to be distilled in a triple effect 
with steam at 228° F. with 10 per cent. loss. Suppose the temperatures of 
boiling are 208°, 188° and 168° and that the temperature of the outlet water 
from the condenser is 85° and the hot well is 90° F. The equations above 
then become 

0.9 [196.5 + 959.4] - 33.1 = ei(972.2) + e 2 (176.2 - 156.1) 

+ e 3 (176.2 - 58.1) 

= ei[0.9 X 972.2 - 0.1 X 176.2] - e 2 [156.1 + 984.7 - 176.2] 

+ e 3 [176.2 - 156.1] 

= e 2 [0.9 X 984.7 - 0.1 X 156.1] - e 3 [136.0 + 996.7 - 156.1] 

1007.3 = 972.2ei + 20.1e 2 + 118. le 8 
= 857et - 964.6e 2 + 20.1e 3 
= 870.7e 2 - 976.6e 3 
e 2 = 1.12e 8 = 0.84 
ei = i.21e 8 = 0.93 
e 3 = 0.75 

The weight of condensate = 3.52 lbs. 

The amount of water used in the condenser per pound of steam in the 
first evaporation is 

M = 0-75I136.0 + 996.7 -58.H _ ^ ^ 

The amounts of heat can now be computed for each evaporator when the 

M 
total amount of evaporation is known. If this amount is M, ^ is the 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 353 

amount of steam condensed in the first stage, ~o M is condensed in the 

second stage, ^9 M is used on the third and ^9 M is condensed in the con- 
denser. By using the methods of this chapter or Chapter III, the surface 
required for this heat under the conditions given can be determined. 

The loss of heat from the surface of the evaporators can be 
computed by the methods of Chapter III, when the actual size 
and shape is known with the temperatures. This would give 
a definite number of heat units instead of the percentage. Its 
use would be the same as above. 

At times the supply for each evaporator is taken from the 
weak liquor line and then the terms 

O2 + e 3 ) q' 2 
e s q f 3 

will be omitted from the equation. Each stage will have the 
weight evaporated in that stage as the unknown in the equation. 
The feed might be heated by the outgoing condensate or if not 
an assumption of its temperature would be made. 



56.54 lbs. Condenser 




Fig. 169. — Hodge's multiple still. 

To remove the air which may collect in the various condensing 
chambers, air pumps must be attached or small vent holes 
must be made from the condensing chamber to the boiling 
chamber. 

Another multiple effect known as Hodge's Multiple Still is 
shown in Fig. 169. In this A is the boiler and B, C, D, E, F, G 
and H are evaporators or stills. Steam from the boiler is 
passed into still B to evaporate its amount M of the liquid 
and enough steam is added to the steam exhausted from one 
still from the boiler steam to give sufficient heat to condense 
the same quantity M in the next still and to make up 
for the radiation and the steam lost by the blow-off from 
the valves K. The blow-off is intended to remove the air from 

23 



354 HEAT ENGINEERING 

the evaporator. / is a condenser and J is a feed-water heater 
to reduce the temperature of the condensate. The feed to each 
still is controlled by a float valve and this feed together with 
the boiler feed is carried through the same line. Since the 
water used in the condenser is sufficient to feed the boiler and 
stills, its amount is limited. Since in addition to this the maxi- 
mum possible temperature leaving the condenser is fixed by the 
pressure of the steam to be condensed, the temperature entering 
the condenser is fixed and hence this water might not be suffi- 
cient to cool the condensate in the feed-water heater to a low 
value. The extra cooling water is required to reduce the tem- 
perature of the condensate still lower. Although not used in 
the Hodge's Multiple Still, a feed heater on the outlet of the 
condensate from each still would increase the efficiency if the 
feed to that still were passed through it. 

To design such a still the desired amount of distillation, or 
the capacity, and the pressures and temperatures would be 
assumed. The total capacity would be divided equally among 
the stills and the condenser; in the above figure among eight units. 
From the amount of water to be evaporated or condensed, to- 
gether with the temperature difference, the probable size of each 
unit is found. From the size and temperature the amount of 
heat lost may be computed. The problems are now solvable 
since equations for each still may be written out as soon as the 
amount of blow is assumed. 

PROBLEM 

Suppose 400 lbs. of distilled water are desired per hour and from the size 
of still required to condense y 8 X 400 lb. or 50 lbs. the radiation is 4000 
B.t.u. per hour and that 3 A lb. of steam blown per hour would be sufficient 
to keep the evaporator clear of air. 

The temperature drop in each unit will be taken as 10° and since the 
pressure even in the condenser is to be above the atmosphere its temperature 
must be above 212° F. Suppose this is taken as 220° F., then the tempera- 
tures of the various stills will be 220°, 230°, 240°, 250°, 260°, 270° and 280°. 
The temperature of the boiler steam will be 290° F. The absolute pressure 
will therefore be 17.2 lbs., 20.8 lbs., 25.0 lbs., 29.8 lbs., 35.4 lbs., 41.8 lbs., 
49.2 lbs. and 57.5 lbs. 

The amount of steam and water leaving the apparatus is 

8 X 50 + 8 X U = 406 lbs. 

This must be the total feed. 

The temperature of the condensing water leaving the condenser would be 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 355 

210° F. if the same temperature difference is to be used here. The feed water 
entering any still is at this temperature. 
The equation for the condenser is 

406g\ + 50%(g' 22 o + r 22 o) = H(q r 220 + r 220 ) -f 50g' 220 + 406?' 2 i 

+ 4000 (33) 
q'i = 69.0 
U = 101° 
The equation for H is 

(50% + M h )(g' 230 + r 230 ) + M f q' 210 = 50g' 23 o + 50U(q' iM + r 22 o) 

+ U(q' M + r 2 3o) + 4000 (34) 



50% + M b + M f = 50% + 50 + % 


(35) 


M b + M f = 50% 


(36) 


M b = 5.79 


(37) 


M f = 44.96 


(38) 


The equation for G will be 




6.54 + M b ) (q' 2 4 o + r 240 ) + M f q' 210 = 50q' 2 4o + 56.54(2' 230 + r 230 ) 




+ %(g' 24 o + r 240 ) +4000 


(39) 


56.54 + M 6 + M f = 50 + 56.54 + 3,4 




M& + M f = 503/4 




M 6 = 6.41 




Mf = 44.34 




The amounts for the stages are as follows: 




Mb M f 




Staged 8.44 42.31 




Stage C 8.03 43.00 




Stage D 7.48 42.98 




Staged 7.18 43.48 




Stage F 6.58 44.11 




Stage G 6.19 44.70 




Stage # 5.79 44.96 




Condenser 50.75 




Feed to boiler 100.44 Feed to tank 305 . 56 




Total feed = 406 lbs. 





The amount of evaporation per pound of steam is TnrTTi = 4.05. A 

result that is only a little better than could be obtained with an ordinary 
quintuple effect. 

The temperature of the distillate is found by adding together the heats of 
the liquid for each discharge and then dividing this by eight to get the aver- 
age heat of the liquid. This is true because each evaporator sends 50 lbs. 
into the distillate line. This gives 

q' m = 223.9 
t m = 255°.l 



356 HEAT ENGINEERING 

The temperature of the water entering the condenser will have to be 101 c 
so that the heat necessary to raise this water from 65° F. will reduce the tem- 
perature of the distillate 

406(101 - 65) 



400 



= 36.3° F. 



or to 218.8° F. To cool this off still more cold water is circulated in the 
cooler. If the distillate is desired to be cooled to 80° F. the amount of 
water needed would be 

% , 400 [218.8 - 80] _„ 

M = ^ — -^ ■ = 3690 lbs. 

oO — bo 

This heat could be saved by arranging the distillate to warm the feed at 
each still. 

DOUBLE BOTTOMS 

The design of double bottoms for evaporation of liquors is 
carried out in the manner of Chapter III. The temperature of 
boiling of the liquor may be fixed by the pressure and found from 
a table or if the temperature is given the pressure may be found 
in tables. If the tables are not at hand recourse may be had to 
the rule of Diihring as given by Hausbrand: 

TEMPERATURE OF BOILING 

The difference between the boiling temperatures of a liquid at 
any two pressures is equal to a constant multiplied by the dif- 
ference in temperatures of water at the same pressures. The 
values of the constants are given for some substances in the 
table below: 



Water 


1.0 


Alcohol 


0.904 


Ether 


1.0 


Acetic acid 


1.164 


Benzene 


1.125 


Turpentine 


1.329 


Mercury 


2.0 


Carbolic acid 


1.20 



If one temperature and pressure is known for a substance 
another may be found. Thus if mercury boils at 674° F. at at- 
mospheric pressure it is desired to find at what pressure it would 
boil at 400° F. Water at atmospheric pressure boils at 212° F. 
The temperature for water corresponding to mercury at 400° F, 
is given by: 

674 - 400 = 2 (212 - t x ) 



CONDENSERS, COOLING TOWERS AND EVAPORATORS 357 

Since the constant in the table is 2. 

t x = 75°. 
Pressure = 0.42 lbs. per sq. in 

Having the temperature required, the temperature difference 
may be found or assumed and then if the heat of vaporization 
is found by reference to tables the area may be found by 

F = KAT (40) 

Where k = 1600 for water. 

k = 1200 for thin liquors. 

k = 500-900 for thick liquors. 

TOPICS 

Topic 1. — What is the purpose of the dry air pump? Why is the air for 
this pump brought into contact with the coldest water? Explain clearly. 
On what does the amount of air depend ? What are the peculiar features of 
the Wheeler dry tube condenser, the uniflux and the contraflo condenser? 
What is the reason for the use of the barometric condenser? 

Topic 2. — Outline the method of design for all parts of a condenser. 

Topic 3. — Sketch and explain action of a cooling tower and derive equation 
for the determination of the amount of water cooled per cubic foot of air 
taken in. Give the expressions for computing each term of the equation. 

Topic 4. — Explain the action of spray nozzles, ponds, and accumulators 
and show how to find size of each to perform a given service. 

Topic 5. — Sketch and explain action of a triple effect. Write formulae for 
the determination of the evaporation per pound of steam supplied to the 
first effect. 

Topic 6. — Sketch and explain action of Hodge's multiple still. Write 
equations for the condenser and the next two stills by which the quantity of 
steam may be found. 

Topic 7. — Explain how to find the temperature at which a substance will 
boil under a certain pressure if the temperature is known for a given pressure. 
What are double bottoms? How are they designed? 

PROBLEMS 

Problem 1. — An air pump is used with a 10,000-kw. turbine. The tem- 
perature of the hot well is 65° F. The vacuum carried is 29 in. How much 
air is present? If the air and vapor are carried around cold water pipes so 
that the temperature is reduced to 50° F., how much has the volume of air 
been reduced? Is the air leakage in this condenser system excessive? 

Problem 2. — Find the amount of surface to use with a 10,000-kw. turbine 
with a steam consumption of 13 lbs. of steam with a vacuum of 29 in. and 
temperature of steam of 65° F. if the cooling water operates from 45° F. to 



358 HEAT ENGINEERING 

60° F. How much water is required? How many %-in. tubes would be 
used in a nest? 

Problem 3. — Find the amount of air to cool 3,900,000 lbs. of water per hour 
from 110° F. to 80° F. in 70° F. weather with the barometer at 29.6 in. and 
the wet bulb at 60° F. 

Problem 4. — How many spray nozzles would be required for Problem 3? 
How large a cooling pond would be required? 

Problem 5. — Compute the amount of distilled water made per pound of 
steam from a triple effect heated by exhaust steam at 3 lbs. gauge pressure if 
the condensed steam in the first effect is of no value, x of the entering ex- 
haust steam is 0.90. Temperature of water supply is 60° F. Vacuum 
allowed on condenser, 20 in. 

Problem 6. — Find the pressure at which mercury will boil at 300° F. 

Find the area required for a double bottom to drive off 700 lbs. of water per 
hour from a solution at 15 in. vacuum. Neglect heat of solution. 



CHAPTER IX 
INTERNAL COMBUSTION ENGINES AND COMBUSTION 

The internal combustion engine has been greatly improved 
within the last 25 years although its history extends back 
several centuries. In 1680 Huygens, a Dutch physicist, pro- 
posed to use the explosion of gun powder to drive the piston 
and in 1690 Papin continued this work. There is nothing definite 
known of their work or similar proposals by Abbe Hautefeuille 
at about this date. In 1794 a patent was granted to Robert 
Street for a gas engine using turpentine in the bottom of a 
heated cylinder and the mixture of this with air was ignited by 
a flame. Samuel Brown originated an engine in 1823-6 in which 
a soluble gas behind a piston was dissolved in water and thus 
produced a vacuum sucking the piston downward. The Wright 
engine in 1833 exploded a mixture of compressed gas and air 
which was admitted to the cylinder after the exploded gases 
were driven out. This engine required two revolutions, or 
four strokes to the cycle. The Barnett engine of 1837 used a 
separate pump to compress the air. In the years 1838 to 1854 
there were eleven English patents for gas engines applied for. 
In 1855 the method of igniting the charge by compressing it 
into a heated tube was patented by A. V. Newton and in 1857 
Barsanti and Matteucci invented an engine with a free piston 
in which the explosion of gas drove the piston upward and the 
contraction due to cooling produced a reduction of pressure so 
that the piston was drawn back by this and its weight, and 
acted on the crank shaft through a ratchet. 

The history of successful gas engines begins with 1860 when 
Lenoir built his machine. This resembled a steam engine in 
structure and action. A mixture of gas and air was drawn into 
a cylinder by the movement of a piston from the fly wheel and 
after the piston had reached the middle of its stroke the gas 
and air were shut off and the mixture was exploded by a high- 
tension spark between two platinum points. The pressure 
immediately rose and as the piston moved forward this gas ex- 
panded reaching atmospheric pressure at the end of the stroke. 
On the return stroke, the same thing occurred on the other side 

359 



360 



HEAT ENGINEERING 



of the piston, the exploded gases on the first side being ex- 
hausted. The cycle of this engine is shown by the card of 
Fig. 170. 

In 1863 M. Alph. Beau de Rochas published a pamphlet in 
which he suggested that the greatest economy of the gas engine 
would be obtained if: 

1st. The greatest possible cylinder volume with the least 
possible surface be used. 

2nd. If the expansion be as rapid as possible. 

3rd. If the expansion be as complete as possible. 

4th. If the explosion pressure be as large as possible. 

He then follows this by a description of a cycle to give these 
results. The cycle consists of: 

1st. Suction on entire stroke. 

2nd. Compression on second or return stroke. 

3rd. Ignition at dead point and expansion on third stroke. 

4th. Exhaust of gases on fourth stroke. 





Fig. 170.- 



-Card of the Lenoir 
cycle. 



Fig. 171.— Card of the Beau de 
Rochas or Otto cycle. 



This cycle was to have ignition due to high compression. 

In 1867 Otto and Langen exhibited their first engine at the 
Paris Exposition. This was similar to that of Barsanti but 
it was a practical machine although very noisy. While Lenoir's 
engine took over 90 cu. ft. of gas per boiler brake-horse-power 
hour this engine consumed only one-half as much. 



OTTO CYCLE 

This was followed by a number of engines by Brayton, Gilles, 
Halliwell, Bisschop, Andrew, Clerk and others, but in 1876 Otto 
brought out the Otto Silent Engine. This engine operated on 
the Beau de Rochas cycle although independently invented by 
Otto. This engine was a great success and the form of cycle 
for many years was known as the Otto cycle. It is one of the 
commonest forms of cycles. It consists, as shown in Fig. 171, 



INTERNAL COMBUSTION ENGINES 



361 



of four strokes: First, a suction stroke ah, at a pressure slightly 
below the atmosphere due to suction of the air from the out- 
side; second, a compression stroke be with explosion at the end 
of the stroke, bringing the pressure from c to d; third, an ex- 
pansion stroke de followed by free expansion ef to a point just 
above the atmosphere; and fourth, a discharge stroke fa at a 
pressure above the atmosphere due to the discharge being driven 
out against atmospheric pressure. 

In the Beau de Rochas or Otto cycle, the lines of compression 
and expansion are practically adiabatics because the action is 
rapid and also because gas is a poor conductor. The gas near the 
cylinder walls is cooled by the water jacket used but this transfer 
of heat probably extends only to this film of gas. The cycle 
requires four strokes for its completion and engines using this 
form are spoken of as four cycle en- 
gines. The cycle is practically 
used on two cycle engines in which 
compressed air is admitted near e 
driving out the burned gases and 
this is followed by compressed gas 
so that when the point / has been 
passed by a small distance, the 
burned gases have been driven 
out and the air and its fuel gas have been introduced. In this 
way there is an explosion for each revolution. 

To study the cycle the lines fa and ab are assumed to coincide 
and eliminate each other giving Fig. 172 as the theoretical cycle. 
This cycle is made up of two adiabatics and two constant vol- 
ume lines. 

The mixture of unburned gases on be and the mixture of burned 
gases on de are different in character and contain different 
amounts of water vapor and carbon dioxide. For this reason, the 
lines are different as carbon dioxide and steam are not perfect 
gases and the variation from the adiabatic of a perfect gas is 
different for the two lines. Moreover the variations of the 
specific heat as the temperature rises also causes changes in 
these lines. 

For simplifying computations so as to study the effects of 
changes on the cycle, a preliminary discussion is often made 
considering the gases present on all four lines to be air and 




Fig. 172.— Otto cycle. 



362 HEAT ENGINEERING 

considering the specific heats as constant quantities. Such 
discussions and results are known as results of the air standard. 

AIR STANDARD EFFICIENCY 

Under these conditions and assuming 1 lb. of substance 
present on the cycle, the heats on the various lines are as 
follows : 

Heat on be = 

Heat on cd = c v (T d - T c ). 

Heat on de = 

Heat on eb = — c v (T e — Tb) 

Work = c v (T d - T c ) - c v (T e - T h ) 

mi i^e C v \l d J- c) Cv\ J- e J- b) /_,>. 

Thermal eff. = rjs = c < Td _ yj (1) 

- 1 _ Te — T b ^ Tb_ Te (t> . 

T d - T c T c T d K * } 



Since 



T c - Tb _ T d - T e 
T c T d 

T e - T b T b T e 



(3) 



T d - T c T c T d 

This is reduced from the cross products of temperature. 

T e T c = T b T d 

This states that the theoretical efficiency of the Otto cycle, 
air standard, is equal to the range of temperature on either 
adiabatic divided by the higher temperature on the adiabatic. 

Now ^ 

But V c is the clearance volume and Vi is equal to the clear 
ance volume plus the displacement, hence 
Vc ID I 



HW 



(5) 

v b ^ "T L)U L T -L 

/ 7 \ ft-l 1 *-l 

Hence 



V b (l + l)D l+l 

I N*- 1 , ■ * 

(6) 
I' 

As / decreases the last term becomes smaller and the efficiency 
increases. That is, the decrease of clearance increases the 



— -<rb)--- to) 



INTERNAL COMBUSTION ENGINES 



363 



efficiency. Of course this increases the pressure at c, hence in- 
creasing the compression increases the efficiency. The amount 
of compression is fixed by the allowable temperature at the end 
of compression and by the pressure at the end of the explosion. 
If the pressure of compression is excessive the temperature may 
be high enough to cause premature explosion while if high with 
a rich gas the explosion pressure is too high. In practice the 
pressure at the end of compression is 80 to 160 lbs. per square 
inch depending on the kind of gas. 
As before: 

T d - T b 



m = 



m = 



*75 = 



a = 



T d 
AW 

V3 



ATKINSQN CYCLE AND DIESEL CYCLE 

Certain other cycles have been proposed for gas engines and 
one will be examined for the purpose of application of theory. 

In Fig. 173 the cycle proposed by Atkinson is shown. In this 
engine pistons were so connected by linkage to the shaft that they 
made strokes of varying length. 
The suction stroke ab is a short 
stroke followed by a compression 
stroke be. The explosion cd is 
followed by a long stroke so that 
the expansion reduces the pressure 
to the initial value. The dis- 
charge stroke ea, which is long, 
brings the pistons to their original points. 

In this case the heat added is 




Fig. 173. — Atkinson cycle. 



Qi = c v (T d 



The heat removed is 



To) 

T b ) 



Q 2 = c p (T e 
Work = Qt - Q 2 = c v [(T d - T c ) - k(T e - T h )\ 

T e - T b 



V-i 



= 1 



k 



T d - T c 



(7) 



364 



HEAT ENGINEERING 



Here there is no simple relation between the temperatures at 
the corners as this is not a simple cycle of polytropics. 

Rudolph Diesel in the last decade of the 19th century pro- 
posed a new cycle of higher efficiency. He recognized the fact 
that to obtain high efficiencies, the cycle of the gas engine must 
approximate the Carnot cycle; that the range of temperature on 
this cycle must be as large as possible and, on account of the loss 
of availability in conduction, the engine must have internal com- 
bustion. He proposed a cycle of isothermal compression from 
a to b, Fig. 174, followed by adiabatic compression be to such a 
high temperature that fuel would ignite on being introduced into 
the cylinder. The fuel was to be introduced at such a rate that 
its burning would produce just enough heat to make the expan- 
sion from c to d isothermal. After the fuel was cut off the ex- 
pansion da became adiabatic. His original paper in the Zeit- 





Fig. 174. — Original Diesel cycle. 



Fig. 175. — Final Diesel cycle. 



schrift des Vereins Deutscher Ingineure was translated in the 
Progressive Age in Dec, 1897 and Jan., 1898. This paper gave 
the results of his work for a number of years. The great pres- 
sure developed by the final compression and the slight gain of 
area by the isothermal ignition led him to abandon the upper part 
of the figure, while the desire to reduce the volume led to cut- 
ting out the lower end of the figure. Thus modified the Diesel 
Cycle took the form shown in Fig. 175, in which adiabatic com- 
pression in a cylinder of small clearance brings the air from a to 
b at which the temperature is high enough to ignite oil which is 
injected into the cylinder. If the burning progresses at the 
proper rate the pressure will be kept constant until cut off at c. 
From c to d adiabatic expansion takes place followed by exhaust 
from d to a and finally to e. The suction stroke from e to a 



INTERNAL COMBUSTION ENGINES 



365 



charges the cylinder with air. This is a four-stroke cycle. The 

efficiency is given by 

c P (T c -T b )-c v (T d -T a ) 
^73 = — 



1 - 



c P (T c -T b ) 

T d -T a 



k(T c -T d ) 
The high theoretical efficiency is due to the high compression. 



(8) 



ACTUAL ENGINES 

The form taken by an actual engine is shown in Fig. 176. 
This represents an engine of the Otto form. In this air is drawn 
into the cylinder by the outward motion of the piston A through 
the valve B. Gas is admitted by C which is so arranged as to 




_J 



_ 



Fig. 176. — Ordinary gas engine. 4-cycle. 

open after B by means of a shoulder formed by a sleeve attached 
to the valve stem. These admission valves are opened by the 
suction of the piston or they may be opened positively by the 
valve gearing. They are closed by the spring D when the piston 
reaches the end of its stroke or when the valve mechanism re- 
leases them. At the end of the expansion stroke the exhaust 
valve E is opened by the lever F operated by a cam. The 
cam is on a cam shaft, operated by a bevel or spiral gear from 
the crank shaft. The speed of this cam shaft for a four-cycle 
engine is one-half the speed of the engine shaft. The governor 



366 



HEAT ENGINEERING 



operates to throttle the mixture or to prevent gas from entering 
as will be explained later. The cylinder K contains a water 
jacket L for the purpose of keeping the temperature of the cyl- 
inder wall low enough for lubrication and also to prevent seizing. 
The two-cycle, Mietz and Weiss oil engine, Fig. 177, is similar 
in action to the four-cycle engine. In this the explosion and 
expansion of the charge compresses air in the crank case K 
and when the piston A overrides the port B the burned gases 
escape to the exhaust pipe M, while at the next moment the 




Fig. 177. — Two-cycle Mietz and Weiss oil engine. 

ports C are uncovered, admitting the air compressed in the 
crank case. This compressed air rushes in and is deflected to the 
head by the projecting finn D so that the burned gases are 
blown out or scavenged. This reduces the air in the crank case 
to atmospheric pressure. After the piston passes C, the air in 
the crank case is rarified, as no air can enter, and after passing 
B the air in the cylinder is compressed. When the piston 
travels back far enough to uncover port E air is drawn into 



INTERNAL COMBUSTION ENGINES 



367 



the crank case from the base of the engine by the partial vacuum 
existing there. In the Mietz and Weiss engine kerosene is 
sprayed into the cylinder near the end of the stroke by the pump 
G, and vaporized by the hot cylinder head H and finally ignited 
by the high temperature from compression in the heated ball 
F at the end. The mixture then explodes and the action de- 
scribed above is repeated. 

In small engines using gasolene, the air entering the crank 
case is drawn through a carbureter in which the air is drawn 
through gasolene or is mixed with it. This charges the air with 
fuel and the mixture is ignited at the end of compression by a 




Scale 120 lb. = l" 
Cylinder Card 



Scale 10 lb. = 1" 
Crank Case Card 



Fig. 178. — Cards from cylinder and crank case of a two-cycle engine. 

spark. In large two-cycle engines the fuel gas and air are 
compressed in piston compressors and admitted to the cylinder 
at the proper time. Fig. 178 illustrates the form of indicator 
cards from the cylinder and crank case of the engine above. 



GOVERNING 

Internal combustion engines are governed in two principal 
ways: (a) by the hit and miss system and (b) by the throttled 
charge system. In the hit and miss method of governing the 
governor acts on the gas or fuel valve so that no gas is admitted 
when the speed exceeds a given limit, but as long as the limit 
is not exceeded the engine receives its full charge. In throttle 
governing the charge of fuel and air may be throttled or the 



368 HEAT ENGINEERING 

fuel may be throttled alone. In the first of these less fuel mix- 
ture is introduced into the cylinder although it is of the proper 
mixture, while in the second case the mixture is changed. In 
each of these the explosion pressure will be reduced and the 
work will be made less. Of course the efficiency in throttle 
governing is reduced. The disadvantage of the hit and miss 
system is the fact that the speed may fluctuate, due to this 
method. In the Diesel engine the governor fixes the point at 
which the oil supply is cut off. 

IGNITION 

The charge in most modern gas engines is ignited by an elec- 
tric spark made by breaking a circuit between two platinum points 
or else causing a high-tension spark to jump a gap between two 
points of platinum or tungsten. The method of compressing 
the charge into hot tube until the mixture comes in contact 
with red hot iron is rarely used at present and compression to a 
high temperature for ignition is used in some small engines. This 
latter method is used exclusively on the Diesel engine. 

The rapidity of ignition depends 
on the pressure and the quality of 
the mixture. Thus with higher pres- 
sures the ignition is more rapid. In 
many cases it is found that the ex- 
plosion is not instantaneous but is 

Fig. 179.— After burning or cont i nue d after the end line as shown 
late explosion. 

in Fig. 179. Such action is called 

after burning. This after burning is found to take place in 
weak mixtures and in throttled supply. The efficiency of the 
engine is found to increase as more air is added with the gas 
than that required theoretically. There are several theories 
given for this, a satisfactory one being that as the heat per 
cubic foot is decreased by the addition of air, the temperature 
increase is not so great and the loss from the cylinder is made 
less, moreover this might give a different specific heat and 
thus, a lower temperature. The after burning may be ex- 
plained by the fact that in many cases the high temperatures 
present prevent chemical combination or may lead to dissociation. 
The amount of dilution by excess air to insure the presence of 
sufficient oxygen is a matter of experiment. The excess air may 




INTERNAL COMBUSTION ENGINES 



369 



be sufficient to make the excess air and inert nitrogen amount 
to about 5 times the gas and the oxygen required to burn it, 
although much larger variations have been observed in explosive 
mixtures. The time of explosion varies with the amount of 
excess air present being slow with large quantities. A slight 
excess gives a quick explosion. It has been found at times that 
the best efficiency was obtained even with 100 per cent, excess air. 

HEAT TRANSFER TO WALLS 

Experiments have been made to determine the loss of heat to 
the cylinder walls in a similar manner to that used for the steam 
engine. Coker found that there was a cyclic variation of 7° C. 
or 12.6° F. at a depth of 0.015 in. in the wall of a cylinder of 
an engine running at 240 r.p.m. In the Seventh Report of the 
Committee on Gaseous Explosion of the British Association, 
the results of Dr. Coker and Mr. Scoble are shown. Fig. 180 is 
taken from this report. 



2000. C 
(3632 E) 



oloOOC 



Oioooc 

g(1832°F) 

a 

Q 

500°C 



Su_ctioji- 



Compression 




Expansion- 



.Discharge 



ISO 



270 300 450 

Angular Fjjame of Crank in Degrees 



540 



Fig- 180. — Cycle in gas temperature from a gas engine after the Gaseous 
Explosion Committee Report. 



This curve shows the temperature of the working medium as 
determined by a platinum couple. The couple was placed in a 
tube which could be cut off from the cylinder by a valve until 
the desired time. The high temperature of the cycle was not 
measured but computed because at this temperature the couple 
would melt if exposed to the gas. These curves show the great 
variation in temperatures which takes place in the gas engine. 

24 



370 HEAT ENGINEERING 

The results of Coker give a variation from 620° F. to 3960° F. for 
one test, while for another it extended from 390° F. to 3250° F. 
The losses from the cylinder walls are due to radiation from 
the high temperature gas as well as from convection currents. 
These losses are affected by the density of the gas. An increased 
density will increase the heat loss. In addition to this fact 
that the loss is made greater by increasing the density through 
the increase of compression, the danger of pre-ignition through 
an increase of the temperature from this cause gives a limit to 
the possible compression. 

FUELS 

The fuels used in internal combustions are inflammable gases 
and oils although Diesel proposed in his patents to use powdered 
solid fuels. The gases in common use are natural gas, illuminat- 
ing gas, producer gas and blast-furnace gas. 

Natural gas is a product of nature found in certain localities 
in various parts of the world, usually where oils are found. It 
is rich in methane. Its heating value is high. T. R. Wey- 
mouth in the Journal of the A.S.M.E., May, 1912, gives the 
following average analysis for natural gas of the United States: 

Methane, CH 4 87 . 00 

Ethane, C 2 H 6 6.50 

Ethylene, C 2 H 4 . 20 

Carbon monoxide, CO 0.20 

Hydrogen, H 2 Trace 

Nitrogen, N 2 5 . 50 

Carbon dioxide, C0 2 0.50 

Helium, He 0.10 

Oxygen, 2 : : Trace 

100.00 

The average heating value of the gas was 887.3 B.t.u. per 
cubic foot at 29.82 in. and 60° F. The specific gravity com- 
pared with air was 0.6135. 

Illuminating gas is only used for small installations as the 
cost is prohibitive. Its heating value is about 700 B.t.u. per 
cubic foot under standard conditions of 760 mm. and 0° C. This 
gas may be the product of distilling bituminous coal, the gas 
amounting to 30 per cent, of the coal. The residue is coke and 
unless it can be sold this method of utilizing coal is not ef- 
ficient. If, however, the illuminating gas is made by enriching 



INTERNAL COMBUSTION ENGINES 371 

producer gas by adding hydrocarbons from crude oil and fixing 
the mixture, this waste of coke would not occur. If the gas 
were made for power purposes alone this enrichment would not 
be made as producer gas is satisfactory for use in gas engines. 
An analysis of illuminating gas is given below: 

Hydrogen, H 2 34.3 

Methane, CH 4 28.8 

Ethylene, C 2 H 4 9.5 

Heavy hydrocarbons, C 2 H 6 1.7 

Carbon dioxide, C0 2 0.2 

Carbon monoxide, CO 10.4 

Oxygen, 2 0.4 

Nitrogen, N 2 14.7 

100.0 

The common form of gas used in power installations is producer 
gas. This gas is made from any form of coal or lignite in a 
producer as shown in Fig. 181 or Fig. 182. These represent two 
forms: the pressure producer and the suction producer. Fig. 181 
illustrates one form of pressure producer of R. D. Wood & Co. 
In this, air is blown into a bed of incandescent fuel A by means 
of the steam jet blower B. The air burns the carbon to C0 2 , 
but on passing through the thick bed of fuel this is reduced to 
CO and the gas rises to the outlet C. The heat of the fire serves 
to drive off the volatile matter from the coal. 

The carbon burning to CO liberates 4450 B.t.u. per pound 
of carbon and this heat would be lost in the scrubber if not 
absorbed in some manner. Some of this heat is used to make 
steam in the top of the producer or in a boiler heated by the 
exhaust gases on their way to the scrubber. If this steam is 
passed into the ingoing air, the dissociation of this by the heat 
of combustion of the fuel will absorb much of this heat. 

The steam used in the air blast cools the gas as it is disso- 
ciated into hydrogen and oxygen on passing over the hot coals. 
This dissociation utilizes part of the heat of combustion of 
carbon to CO and so utilizes some of that which would be lost 
in the scrubber. The steam also prevents the fire from clink- 
ering. The gas now contains CO, H 2 ,0 2 and N 2 as well as some 
volatile gases. 

Fresh coal is discharged from the hopper D into the chamber 
E and from this it is distributed by a special device F which 



372 



HEAT ENGINEERING 



produces in a uniform bed. The openings G in the top are for 
the introduction of slice bars to break up the fuel bed. The 
openings H on the side are for observation of the fuel bed and 
for barring the bed when necessary. The producer is capped 
by a water-cooled top and the sides are lined with fire brick. The 
ashes drop into a water-sealed base. From the producer the 
gas passes through the outlet C into a scrubber which consists 




Fig. 181. — Pressure producer of the R. D. Wood Co. 



of a metal cylinder filled with coke and cooled by water. The 
gas enters a water seal at the bottom of the scrubber and leaves 
at the top. From this point the gas may be taken to a tar 
extractor if the gas contains tar to any extent. This is practi- 
cally a centrifugal fan over which the gas passes, being thrown to 
the outer edge and thus causing the heavier tar to be caught by 
the casing. From this point it is taken to the gas holder. 



INTERNAL COMBUSTION ENGINES 



373 



In Fig. 182 the Otto Suction Gas Producer is shown. In 
this a suction is produced by the engine drawing in gas. This 
draws air from the atmosphere at A over the hot water in the 
top of the producer at B and thence through the pipe R to the 
bottom of the fuel bed C. The water is heated by the gas passing 
from the producer. It then passes through the down pipe D 
to the seal box, depositing tar or dirt at the bottom and passing 
up through the scrubber E; it finally enters the receiver F and 
then passes to the engine. 




Fig. 182. — Suction producer of the Otto Gas Engine Co. 



The blower G is used in starting the fire in the producer. The 
three-way cock H is used to direct the gases and smoke to the 
chimney before burnable gas is produced in starting a fire. The 
scrubber E is filled with coke over which water trickles and 
through which the gas passes. The spherical charging ball 
at the hopper mouth prevents gas discharging or air entering. 

An analysis of producer gas from soft coal gave the following: 



374 HEAT ENGINEERING 

Carbon monoxide, CO 20.9 

Hydrogen, H 2 15.6 

Carbon dioxide, C0 2 9.2 

Oxygen, 2 . 0.0 

Ethylene, C 2 H 4 0.4 

Methane, CH 4 1.9 

Nitrogen, N 2 52.0 

100.00 

This gas gave 156.1 B.t.u. per cubic foot under standard 
conditions. 

Since 1900 blast-furnace gas has been used successfully. 
Apparently it was first used in England in 1895 on a 12 X 30 — 
15-h.p. gas engine and in the same year on an 8-h.p. engine in 
Belgium; in 1898 a 150-h.p. engine was used; in 1899 a 600-h.p. 
engine was used in Belgium and exhibited by Cockerill & Co. 
at the Paris Exposition of 1900. These were followed by larger 
engines. In 1903 the Lackawanna Steel Company installed a 
number of 2000-h.p. engines using blast-furnace gas. In most 
cases the gas is cleaned by centrifugal washers before it is applied 
in the gas engine. This gas was formerly used beneath boilers 
for steam generation but gas engines utilize the heat more 
effectively. 

An analysis of blast-furnace gas is given herewith: 

CO 25.83 

C0 2 9.37 

CH 4 0.54 

H 2 2.96 

N 2 61.30 

100.00 
Heat value = 105 B.t.u. per cubic foot. 

Crude oil is usually used by the Diesel engine. This oil 
varies some in composition and heating value. The analysis 
below is for one form : 

C 84.9 

H 13.7 

O 1.4 

100.0 

Its heating value is 19,000 to 20,000 B.t.u. per pound. 

Gasolene is the light first distillate from crude petroleum oil. 
It is easily vaporized and in gas engines it is usually delivered to 
the air supply by a form of mixer known as a carbureter. Its 



INTERNAL COMBUSTION ENGINES 375 

heating value is about 20,500 B.t.u. per pound when the steam 
formed is reduced to water. It contains approximately: 

C 83 . 5 per cent. 

H 15.5 per cent. 

N 2 , S, and 2 1.0 per cent. 

Kerosene is one of the later distillates of crude mineral oil. 
It is used on a few engines. It requires a special form of car- 
bureter or vaporizer, requiring a high temperature to properly 
volatilize the oil. The analysis of this gives: 

Carbon 84^ per cent. 

Hydrogen 14 per cent. 

Nitrogen and oxygen V/2 per cent. 

The heating value will be about 19,900 B.t.u. for the higher 
heating value. 

COMBUSTION OF FUELS 

The combustion of these various gases is accomplished by the 
mixing of air with the fuel. To get a mixture which will explode 
or burn rapidly the quantity of air must be within certain limits, 
usually the amount necessary for complete combustion plus 15 
per cent, will give good results. To find the weight or volume of 
air to burn any given constituent, reference is made to chemical 
formulae and Avogadro's Law. 

Thus to burn 1 volume of CH 4 , 2 volumes of oxygen are 
required — which means 9.54 volumes of air. From this burn- 
ing 1 volume of C0 2 results and 2 volumes of H 2 0. The water 
may condense. The weights of these per pound of CH 4 are: 
4 lbs. oxygen, 17.40 lbs. air, 2% lbs. C0 2 and 2}i lbs. of 
water. These statements are seen from the following: 

CH 4 + 20 2 = C0 2 + 2H 2 

By Avogadro's Law: 

1 volume CH 4 + 2 volumes 2 = 1 volume C0 2 + 2 volumes 
H 2 0. 
By chemical equivalents: 

[12 + 4.032] lbs. CH 4 + 2[16 X 2] lbs. 2 = [12 + 32] lbs. C0 2 
+ 2[2.016 + 16] lbs. H 2 0. 
or 

1 lb. CH 4 + 4 lbs. 2 = 2% lbs. C0 2 + 2J4 lbs. H 2 0. 



376 HEAT ENGINEERING 

Air contains 77 parts by weight of nitrogen and 23 parts oxygen 
while by volume the proportion is 79 to 21. Hence 2 volumes of 

2 
oxygen mean pr^r or 9.54 volumes of air and 4 lbs. of oxygen 

4 
mean tt-^ = 17.40 lbs. of air. 

The heat of combustion of CH 4 is 21,566 B.t.u. per pound if 
the water produced remains as steam or 24,019 B.t.u. per pound 
if the steam is condensed to water. The former is spoken of 
as the lower heating value, the latter as the higher. 

In gas-engine work in England, Germany and America, the 
lower value is taken in determining the efficiencies, while in 
France the higher value is taken. It is fairer to use the higher 
heating value and charge the engine with the heat which is pres- 
ent with the steam in the exhaust. 

To find the amount of heat per cubic foot of gas the numbers 
above would be divided by the volume of 1 lb. of the gas 
under certain definite conditions. The conditions assumed in 
many cases are 14.7 lbs. per square inch pressure and 32° F. 
although for natural gas the conditions are often a pressure of 
28.7 in. of mercury and 60° F. For the former case the volume 
of 1 lb. of gas is given by: 

1544 
TT MBT MRT 1 l X mol. wt. X 



X 



mol. wt. 14.7 X 144 

359 



mol. wt. 



(9) 



The specific weight is given by: 



1 mol. wt. , N 

m = v = ^59~ ( 10 ) 

359 
For CH 4 : v = 16 032 = 22A cu - ft - 

This gives the values of 1072 B.t.u. per cubic foot for the greater 
heating value and 963 B.t.u. for the lower value for CH 4 . 

Using the methods above for various gases and elements the 
following table is computed : 

GAS COMPOSITION 

To study the action of the gas engine theoretically the actual 
gas supplied and the amount of air or the products of combustion 



INTERNAL COMBUSTION ENGINES 



377 















































ID 


OS 






ID 




*os 




































* S N H h 00 CM W5 


t- rH 
































•' 


lONNNOHNH^OO 


'apixoip 
jnu'dpig 


O CO 
































• 


rHOrHOrHOOOOrH 



















1 
















; ^ «-• 


OOOOOOOOrHO 




































■* 00 <M 1> 


z oo 




































■* rfi 00 >D O tJ* I s -. 


r- 


















! O O 
















ONiOHCNNH O 


'apixorp 


rH 


















! O O 
















CMOrHOrHOrHOCOOO 


uoqjB'3 


■* 00 


















J r-l rH 
















O O d d O O* O O CO O 







































S" ' 




00 CM 

• •* 00 CO CO X 


0'"H 


T-t O 






























O O 




<D 




• l> rH CO rH CD 


O OS 






























O O 




; 




• CO O CM O CO CD 




00 OS 




































• 0" d d co 




.—1 rH 


































































rH CD ID 




































rHOSTHCOrHOO CM ID 


•uaSoj^i^; 


<N O 
























• O O 








TflHSHMHtDHOOOO 


00 
























O O 








CMOrHOCMOrHOCOrH 


X CM 
























rH rH 








O O d 0' O O O* O CM O 




CM i-i 
























O 1-H 










• O O ID CO 
■ O O CO b- 






O O »D CO 

O O co i> 








r}< 00 ID 
00 Q iO M (M 00 CM K> 

T-lrHlCrHOrHT^rHCOOO 


zq 'uaSAxQ 


O <N 
<N rH 










. rH T-l Tj* Tf* 

: 1 1 1 1 






rH rH CO CO 
1 1 1 1 








<NOrHOCNOrHOOrH 

Q O d OO 0' O O CM C5 




























co b- 10 














• CO rH O O 






M H N O) 








OSrHCOTHOOiDCM ID 




O 










• CM CM O O 






(M N N N 








TfHNHNHlOHHOO 


J TV 


00 ■* 
























CMOrHOCMOrHOCOrH 












• O O rH rH 






OOOO 












00 CM 










' 1 1 1 1 






1 1 1 1 








oddodooo'cMo' 




(N rH 






. fill 




























CS| CM ID 


S*H 


co 


















IDCOCM00 OOrHCM >D 


00 <N 





• os 


■ H O Tf IO 










CO »D CO O 00 O 


TtHCMOSrHOSrHCOrHlOOO 


'apiqdjns 


O lO 


CD 




• Tj< ID rH rH 










t> CD ID O 00 O 


CMOrHOrHOrHOOSrH 


uaSojp^jj 


CO —1 


CO 


• CO 


• H H © N 










T|i IOOH H H 


O O d O" O O O O rH O 




1> 







. O -ID • 










tH • • • 




S ' Z 0S <n 


O • 










• O 


CO • 










CO 






• 




















jnqd^g 


CM 

co • 






: ^ 


^ • 










co 






CM • 








































O CO U3 


OO 






















MOlMMHOOON ID 


O O 


10 


t^ 


•N O O) 00 


b- O 




(N 00 








1<HNHrtH!OHC000 


'spixouour 


O 00 


CM 


• co 


- ID >D Tt< CO 


ID O 




Gi 00 








CMOrHOCMOrHOCOrH 


uoqjBQ 


00 (N 


CO 


CO 


• O O CM CM 


rH r-< 




l-I rH* 








d 0' O O O 0' 0' O <N O 




CM rH 1 "^ 




















O • 


Q 




co • cs ■ 


CO 




co • 










O'OO 


O • 


LO 






. CO 


1> • 


CO 






^ 




























04 uoqj'CQ 


CM • 






rH • 


lO • 


<M 






•* 






























O • 


■<* 






t> • 


O • 


l> 






co ■ 




























0'*00 


O • 


<* 






CO • 


CO ■ 


CD 






C5 




























0; uoq^B^ 


CM ■ 






CM • 


rH ! 


CO 






00 












































































ID 




00 


OO1D1DCOOO1D 
















ID 00 OS <N ID 


9 H z O 


■* b- 


OOC000t>>DCMcD 


CO O 




t^ ID O O 










rH rH Tt< rH CM 00 


O OS 


O CM 00 CO • • • • 


02 




TH rH 00 O 










CM O rH O CM rH 


•auBq^ 


rH 


nohh wm SS 


CM CM 




<N CO rH CO 










d O CM d 




CO H 


CM CM 






rH rH 




















CD >D ID 




CM 


OOOiOCOOOO 










TtHi-HOCDrHOOOCM ID 


>H z O 


CO 


OONtC^OOrt 


■<# O 




t> O 00 O 




OM*NMH«HN00 


00 


■^ O CD >D • • • ■ 


rH O 




■* CO CM O 




^OCOOCMOrHOCOrH 




^ Q H H ^ ^ ^ ^ 
CN CM rH rH 












00 CM 
CM rH 


CO CM 




rH rH rH CM 




d O O d O O O O CM d 
















ID rH ID 




CM 


OOrHCOOOO-tf 










COCDOO<N)OOOSCM 10 


>HO 


CO O 


OOt^-^OO-^LO 


ID O 




O T)l lO O 




OINKJNOHNHCIOO 


O •"# 


O ^ O O) • • • • 


b- O 




■* ID CM O 




lOOiOillONOHH 


'au'Bqjaj^ 




rji" i-T t-T ^ « N d 














CD CM 
rH CM 


CM CM ""* 


CM rH 




CO i t^ CM CM* 




dodooddo"*d 




CD 












CM 00 ID 






OO»DrHOOO00 














tH OS CD CO 00 CM ID 


Z H 
'uaSojpXjj 


O O 


OOi'ffiOiONW 










O 00 O O • 




(MrH-H^rHrHrHCMrH 00 




ID 00 CO CM • • • • 










i> 00 • 




rHOr^OCMOCMOOrH 


CM 00 


- - 00 O tH CM 
















t^ 


CD ID W 










CO r-i OS rH • 

CM 




COOCMOCOOCMOCOO 
CO 
















. -P . £ 




.p 


^ 


+3 


+3 


^j 




































<*-> | 


«« • 






«tH 


































: 3 : 3 




3 : 


3 


3 


3 : 


3 


































.5.0 
















u 


































i3 S^ » 






J '■ 


J 




^ 




_^ 




_^ 
















- 1 a rt ft 


~ a— ft ~ a . a~ a 




























(-. ti 


u U 1-> u 1-* 


























<U ^i 0) +a 


<U -p CO -ti 0>4J 8 <j « jj 










Ji ^ 


-D . 


ft <~ a «« 


a «« a^ a •« n^ a«~ 






-Q 


— O — ' O £ 3 .O 3 


JD 3 XJ 3 .D 3.D 3 J2 3 






a ft 
« £ 


M jWq JO JO 


qOjO^O^O^O 


P-fpHPLlPHrHPHrHrHP-lP-l 


B.t.u... . . 
per lb .. . 

B.t.u 

per cu. ft. 
Oxygen. .. 
required... 
Air 
required... 


8 £ g 


J5 

<? c? e •« 


ur 


pi 


nc 


TTI 


DO 


J° 


B1 


an 


po 


Id 



378 



HEAT ENGINEERING 



must be known. From these the various temperatures and forms 
of lines may be determined. From the chemical analysis of the 
fuel gas and the exhaust gas, the amount of air may be found, as 
will be shown below. 

Suppose then that the producer gas given on p. 374 is to 
be burned in a gas engine with 25 per cent, dilution and it is re- 
quired to know the amount of air to be admitted, the volume of 
the mixture, the heat per cubic foot of mixture and the products 
of combustion. 





Per cent. 
Vol. gas 


Volume 
air 


Heat of 
combus- 
tion 


Products of combustion 




CO2 


2 


N 2 


H 2 


SO2 


CO 


20.9 
15.6 
9.2 
0.0 
0.4 
1.9 
52.0 


49.9 
37.2 


7,060 
5,130 


20.9 




39.3 . 






H 2 ..... 


29.5 


15.6 . 




C0 2 . . . 


9.2 








2 


0.0 

5.7 
18.1 














669 
2,037 


0.8 
1.9 




4.52 
14.30 
52.00 

21.30. 


0.8 . 




CH 4 ... 


3.8 . 




N 2 














5.82 






Excess 


100.0 
Air .... 


110.9 

27.7 


i4,896 


















138.6 


14,896 


32.8 


5.82 


160.92 


20.2 





This table has been prepared from table on p. 377 for 100 lbs. 
of gas in the following manner: 

Air for CO = 20.9 X 2.38 = 49.9 
High Heat for CO = 2.09 X 337 = 7060 
C0 2 from CO = 20.9 X 1.0 = 20.9 
N 2 from CO = 20.9 X 1.88 = 39.3 

B.t.u. per cubic foot mixture = ^90 ^ = 02.5 

1 ^8 6 
Air per cu. ft. in0 = 1.386 

Vol. of mixture = 2.38 cu. ft. 

Products of combustion per cubic foot: 

C0 2 = 0.328 cu. ft. 
2 = 0.058 cu. ft. 
N 2 =1.609 cu. ft. 

The values of B, c p and c v for a mixture are important to con- 



INTERNAL COMBUSTION ENGINES 379 

sider. Since the molecular weights of the various constituents 
are different from each other and since the mixture contains CO2 
and water vapor the expressions for specific heat must be com- 
puted. The a's of the expression 

c = a + bt 

are different for these various substances and the same is true 
for a"s and 6's. If the values of a, b or a/ for 1 lb. of a gas be di- 
vided by the volume of 1 lb. under standard conditions the 
value of these quantities for the heat to be added to 1 cu. ft. of a 
substance to change its temperature 1 deg. will be found. This 
might be called the specific heat of 1 cu. ft. The reason for using 
this quantity is the fact that the gas analysis given is usually by 
volume and the necessity of reducing this to percentage by 
weight is eliminated. This is done with the heat of combustion 
to reduce that to the heat per cubic foot. 

If v p is the percentage volume or the relative volume of the 
various gases in a mixture the following equations hold for the 
mixture : 

v p = partial volume of any constituent. 

Mol. wt. mix = — — (11) 



(12) 
(13) 
(14) 

(15) 

'(16) 
(17) 

(18) 
(19) 
(20) 



Mb. mix 


VVp 


("mix ~ 


?{vp X a) 
Zv p 


Omix 


2(v p X b) 
2v p 


& mix == 


?{vp X a') 

2v p 


tl mix == 


?{v P X H) 
Zv p 


c — 

^ pmix 


("mix \ Omix-L • 


^ v mix ~ 


d mix ~l Omix-L • 


■D-mix = 


1544 


Mol.Wt.mix 




359 B 


V = 


Molwt. ~ 4.3' 


k = 


^ pmix 
t-' vmix 



380 



HEAT ENGINEERING 



PROBLEM 



Suppose the natural gas given on p. 370 was used in a gas engine and 
it is required to find the amount of air used per cubic foot of gas and the 
true products of combustion if the exhaust gas analysis by volume is : 

C0 2 = 8.8 per cent. 
N 2 = 85.7 per cent. 
O2 = 5.5 per cent. 

These analyses show no water vapor from the burning of the gas nor from 
that taken in with the air and gas. Suppose the gas is at 80° and saturated 
while the air is taken at 70° and is H saturated. 

The pressure of the water vapor in the gas is 0.5056 lb. per square inch, 
while that in the air is 0.1814 lb. per square inch. The barometer is 14.7 
lbs. per square inch. Now from Dalton's Law the moisture percentage 
by volume is equal to the percentage of the dry air pressure which gives 
the partial pressure. Hence 

Per cent, volume of gas entering equal to moisture = 

0.5056 



14.7-0.5056 



3.56 per cent. 



Per cent, volume of air entering equal to moisture = 

14.7 - 0.1814 " iZb per cent - 

On account of pressure and temperature differences of gas and air the 
theoretic amount of air per cubic foot must be multiplied by 

14.7-0.5056 460 + 70 

14.7 - 0.1814 A 460 + 80 u ' y 

to give the actual cubic feet of air per cubic foot of gas. 

The true constituents of the gas are those given on p. 370 multiplied by 
0.964 to which 3.56 per cent, moisture is added. These quantities are then 
multiplied by the quantities of the table on p. 377 to give the various 
volumes of the products of combustion and the volumes of the air required. 
The results are expressed as a percentage. If the total air required is 
multiplied by 1.25 per cent, the moisture from the air is found. 

The computation is now made. 





Per cent, 
volume 


CO2 


N 2 


H 2 


Air 


Heat 


CH 4 87.00X0.964 

C2H4 

C2Hfi 

CO 

N 2 + He 


83.90 
0.19 
6.23 
0.19 
5.41 
0.48 
3.60 


83.90 
0.38 

12.46 
0.19 

0.48 


632.0 

2.15 

82.50 

0.35 

5.41 


167.80 

0.38 

18.69 


800.00 
2.72 

104.25 
0.45 


90,100 

318 

11,420 

64 


C0 2 








H 2 




3.60 


















100.00 


97.41 


722.41 


190.47 


907.42 


101,902 



INTERNAL COMBUSTION ENGINES 381 

Moisture from air = 907.42 X 0.012 = 10.90 
Total moisture = 201.37 

The volume of the mixture before burning is 

100 + 907.42 + 10.9 = 1018.32 

while after burning the volume is 

97.41 + 722.41 + 201.37 = 1021.19 

There is a slight increase in volume. 

Products of combustion in the original analysis show that there is some air 
present. The amount of oxygen is 5.5 per cent. This is associated with 

5 5 
Triyr X 0.79 = 20.6 per cent, of nitrogen. 

The nitrogen in the analysis from combustion must be the difference 
between 85.7 per cent, and 20.6 per cent. 

Nitrogen from combustion = 85.7 — 20.6 = 65.1 per cent. The nitro- 
gen in the products of combustion give a total of 722.14 parts of which 5.41 
have been brought in by the fuel gas, or 

7^X4 X 65A - °' 49 

is the amount of nitrogen chargeable to gas. The nitrogen from the air 
required to burn the gas is given by : 

65.1 — 0.49 = 64.61 per cent. = volume of nitrogen 

from air to burn gases. 

Of IJ 

The total nitrogen is ~^r times the nitrogen from combustion, or. 

85 7 

t^t X 722.41 = 952 

oo.l 

The free air is ^tj X 907.4 = 289.6. 

Total air is 907.4 + 289.6 = 1197. 

Moisture with free air = 0.012 X 289.6 = 3.48. 

The products of combustion are: 

C0 2 97.41 7.42 

N 2 952.00 72.40 

2 |^X97.41 61.10 4.62 

H 2 | 2 °3'4g} 204.85 15.56 

1315.36 100.00 
As a check, the total volume of the mixture before burning is : 

Volume gas 100 . 

Volume air 1197.0 

Volume moisture 14.4 

Total 1311.4 



382 HEAT ENGINEERING 

The theoretical air per cubic foot is 

907.4 n n „ 
-™- = 9.07 cu. ft. 

The theoretical air at atmospheric conditions 

9.07 X 0.96 = 8.71 cu. ft. 
The actual amount is 

1197 X 0.96 An 
=^r = 11.49 cu. ft. 

The heat per cubic foot of mixture entering is 

101902 

T 3 ira = 77.6 B.t.u. 

The specific heat of the gas mixture entering is given by the following : 

Volumes 

Air 1197.00 

CH 4 83.90 

C 2 H 4 0.19 

C 2 H 6 6.23 

CO 0.19 

N 2 + He 5.41 1292.92 

C0 2 0.48 0.48 

/ 3.60 

M2 ° \ 14.40 18.00 

1311.40 
For the total volume of 1311.20 units the following is true: 
Va Vb 

1292.92X0.018 =23.3 1292.92X0.185=239.0 

0.48 X 0.0202 = 0.0098 0.48 X 0.807 = 0.3875 

18.00 X 0.0188 = 0.339 18.00 X 0.668 = 12.0 

23.6488 251.3875 

Ya' 
1292.92 X 0.0125 = 16.2 
0.48 X 0.0147 = 0.007 
18.00 X 0.0132 = 0.238 
16.445 
VC P = 23.6488 + 251.39 X 10~ 5 T 
For one unit volume 

C P = 0.0180 + 0.191 X 10~ 5 T 
VC V = 16.445 + 251.29 X 10" 5 T 
C v = 0.0125 + 0.191 X 10~ 6 T 
For T = 1000° F. these become 
VC P = 26.1588 
VC V = 18.9554 

Cj, _ 26.1588 _ 
k ~ C v ~ 18.9554 ~ 1,d8 



INTERNAL COMBUSTION ENGINES 



383 



The B for this mixture of gas and air is found by (18) 



Volume 

Air 1197.00 

CH 4 83.90 

C 2 H 4 0.19 

C 2 H 6 6.23 

CO 0.19 

N 2 +He 5.41 

C0 2 0.48 

H 2 18.00 





Molecular 




Relative 


Per cent. 




weight 




weight 


weight 


X 


28.8 


= 


34,420 


94.4 


X 


16 . 032 


= 


1,344 


3.7 


X 


28.032 


= 


5 




X 


30.048 


= 


189 


0.5 


X 


28.0 


= 


5 




X 


28.02 


= 


151 


0.4 


X 


44.0 


= 


21 


0.1 


X 


18.016 


= 


324 


0.9 



36,464 



100.00 



Mol. wt. 



B m = 



36464 
1311.40 
1544 

27.78 ~ 



27.78 



55.7 



For the exhaust 

N 2 

2 


gases 


72.40 
4.62 


77.02 


Va 
77.02 X 0.018 = 1.387 


co 2 


7.42 


7.42 


7.42 X 0.0202 = 0.1496 


H 2 0. . . 


15.56 


15.56 


15.56 X 0.0188 = 0.2925 



100.00 



1.8291 



Vb 



Va' 



77.02 X 0.185 = 14.25 

7.42 X 0.807 = 5.99 

15.56 X 0.668 = 10.35 

30.59 



Cr 



= 1.829 + 30.59 X 10" 5 T 
, v = 0.0182 + 0.306 X 10- 5 T 

YorT = 2000°F.;FC P = 2.441 



77.02 X 0.0125 = 0.965 

7.42 X 0.0147 = 0.109 

15.56 X 0.0132 = 0.205 

1.279 

VC V = 1.279 + 30.59 X 10~ 5 T 

C v = 0.0128 + 0.306 X 10~ 5 T 

2.441 

VC V = 1.891; k = = 1.29 

1.891 



By the method used for the mixture, the B for the ignited gas is B = 55.7. 
Although this value of B is the same as that for the unburned mixture, the 
composition is so different that the expressions for specific heats are not the 
same. 

If mixtures on the compression and expansion are those assumed, the 
values of the various quantities are as follows : 

For compression: 

C pm = 0.0180 + 0.191 X 10- 5 T 
C vm = 0.0125 + 0.191 X 10~ 5 T 
For 1000° F., a mean temperature: 
C p = 0.0199 
C, = 0.0144 



384 



HEAT ENGINEERING 



_ 0,0199 _ 
* 0.0144 ~ 1,d8 
B m = 55.7 

# = 77.6 B.t.u. per cu. ft. 
v = 12.95 

For the exhaust gases : 

C pm = 0.0181 + 0.295 X 10" 5 T 
C vm = 0.0128 + 0.295 X lO" 5 T 

For T = 2000° F. : 

C vm = 0.0239 
Cvm = 0.0186 
0.0239 
0.0186 
B m = 55.7 
w = 12.95 



ft = 



1.283 



TEMPERATURES AT CORNERS OF CARD 

The various temperatures at the corners of the indicator card 
of the gas engine cycle are now computed theoretically. In 
this computation the variation of specific heat will be disregarded 
at first after which the effect of this variation will be investigated. 
In a gas engine the exhaust gas which remains in the cylinder 
warms the incoming gas and changes its temperature. For 

this reason the amount of gas 
actually used by an engine cannot 
be told by the change in volume. 
Moreover the heat removed by 
the cylinder jacket during the 
different events affects the results. 
Consider the card of Fig. 183. On 
the suction stroke 5-1 a charge of 
gas and air is drawn in, mixing 
with the burned gases of volume 
7 5 in the clearance space. The 
temperature of the gases in the clearance space T 5 will be equal 
to the temperature resulting from the expulsion of the exhaust 
gases from point 4. The gas which remains in the cylinder at 4 
acts on the gas driven out to force it into the atmosphere and 
hence it may be considered to expand adiabatically in driv- 
ing out the exhaust. The gas in contact with the piston from 
1 to 5 may be considered to be at a constant temperature equal 




Fi 



183. — Theoretic form of in- 
dicator card. 



INTERNAL COMBUSTION ENGINES 385 

to that due to adiabatic expansion from 4 to 1 if no loss or gain 
of heat from the cylinder walls be considered. Thus: 

T s = T t (f)^ (21) 

But * = ^ = |? (22) 

Pi Ps T z 



Hence 



>(y)~^ w 



The mixture of this burned gas of weight 

Mo BTb 

and the fresh charge will result in a volume of gas Vi at a pressure 
Pi and at a temperature TV The weight of this mixture of 
fresh air and gas and the burned gas will weigh M 2 pounds. 

Although the values of B are not quite the same in these two 
formulae they may be considered the same in this work. 

If this gas enters the cylinder from 5 to 1 the work done by 
the entering gas on the piston, pi(Vi — Vz), will just equal the 
work done by the atmosphere in forcing the air into the cylinder 
so that this need not be considered in equating the energy at 5 
plus the energy in the air entering to the energy at 1. 

£*J + (M 2 - M ) Jc v T a = ^J (24) 

Ph = Pi. 

Pi(Vi ~ Vs) (PiVi igsTtX 

0.4 "~ XBT, BTj JCvla 

Pi(Vi- Vi _ Pi (Vi _ Vj\ ... 

0AT a "OiVT 7 ! TJ K J 

Now the clearance V 2 or V& is given as I times the displace- 
ment of the piston. 

V 5 = l[V 1 - V 6 ] (26) 

F 5 = ji^ F, (27) 



Hence -*- Yl = (l±l - ±)-Jl- 

l + lT a \ Ti Tjl + l 



25 



386 



HEAT ENGINEERING 
^ (1 + Q(T a T 5 ) 



T 6 + IT a 

Substituting for T b its value from (23), T\ reduces to 

•T*\ *— J- 



(28) 



T x = 



( i + 4^ 4 y— ] 



T,y)— +it. 
■-C+V"' 



■T, 



T* = — 



(29) 

(30) 
(3D 
(32) 



Now the heat added from 2 to 3 is called VH and is equal to 
VC V (T 3 - Ti) hence 



T * - T * + VUl ""* Tl + g- 



Hence 



n = T 7 ! + 



Cv'OL 



(33) 
(34) 



These equations reduce equation (29) to 

H 



( 1 + ^+cd 



T, = 



« + 



# 



C V T\ 



\™-& 



a + 



H 



C V T, 



k-i 
" +IT C 



(35) 



In this equation the only unknown is T\ but the equation is 
implicit so that its solution is best made by trial. After 7\ 
is found the other temperatures may be found in succession. 
Thus suppose in an engine with 25 per cent, clearance, the out- 
side gas is at a temperature of 70° F. and the heat per cubic 
foot is 80 B.t.u. while the value of C v per cubic foot is 0.013. 



1.25\ - 4 



a = l025/ = L 



9 



INTERNAL COMBUSTION ENGINES 



387 



T, = 



1.25 X 530[?\ + 


80 1 
0.013 X 1.9J 


1.9- 


1.9 

80 
1 0.013 X T l 


k-i 
k 


L ri+ 0.013 X1.9J 


1.9 

19| 8 ° 

L ' 0.013 x rj 


k-i 
* +(0.25X530) 



Since the last term in the denominator is small the value of 
Ti is practically equal to (1 + l)T a although on account of 
another term being additive, the value will be slightly smaller. 
(1 + l)T a = 1.25 X 530 = 662. Hence try T x = 625. 



1.25 X 530 X 3855 X 0.593 
I 1 = oo^r- . , ^ ^o — ; — TW?i = y>2>c> 



If 627 is tried 



3855 X 0.593 + 132 



1.25 X 530 X 3857 X 0.618 



Use 626. 



1 ~ 3857 X 0.618 + 132 
T 2 = 626 X 1.9 = 1190 

80 



626 



T 3 = 1190 + 

7340 
U _ T9"~ 



0.013 
3860 



= 7340 



These are all higher than the values found in practice due to 
the fact that the combustion on the line 2-3 is not always 
complete at the point 3, that the jacket removes heat from the 
cylinder and also because the specific heat varies with the tem- 
perature increasing perceptibly at the higher ternperatures. 
If the heat of combustion from 2-3 be reduced to 75 per cent, 
of its value the temperature T 3 will be materially decreased. 



ADIABATICS 

The compression and expansion lines of the card have been 
assumed to be adiabatics of the form 



pv 1A = const. 



(26) 



This assumes that the gases are perfect gases and that c p and 
c v are constant. Now the true forms for the specific heat are 



and 



c p = a + bT 
c v = a' + bT 



(16) 
(17) 



388 HEAT ENGINEERING 

and the equation of the adiabatic is 

= c v dt + Apdv = (a + bT)dt + Apdv 

% + Mt=- AB^ (36) 

T v 

In the above expressions the quantities refer to 1 lb. al- 
though as will be seen later it might be simpler to have them 
refer to 1 cu. ft. 

The integration of this between limits finally gives 

a' log e ~ + b ( T * ~ TO = ~ AB loge y x ( 37 ) 

7 2 
From this T 2 may be found if T\ and ^- are known. 

, V* , T 2 V 2 

log - log ^r - log =- 

Now n = £ = * T/ Vl (38) 

l0 § F 2 l0 S F 2 



Since 



Pi !Ti 7 2 
In a form involving T2 and T\ only, this may be written 

ABlog e p + a'logep + b(r 2 -r 1 ) 
a' loge ^ + b(T 2 -T,) 

Of course since — must be known to find T 2 and Ti this form 

is not necessary, except in cases in which it is desired to know 
the value of n over a given range of temperatures. This is a 

closer value of n for the range from T 2 to T\ than the value — • 

Equation (37) may be written 

a' log e T + bT + AB log e 7 = const. (40) 

a' loge 7? + a' log e 7+ A5 log e 7 + bT = const, 
a' log e p + a log e 7+ 6T = const. 

p a'ya e bT _. ^a/ya e ^g~ = COnst. (41) 

These equations considering the variation of c„ with tempera- 
ture will cut down the pressure of compression and its tempera- 
ture and thus affect the expression for efficiency of the air cycle. 



INTERNAL COMBUSTION ENGINES 389 



EFFICIENCY CHANGE 

The change in efficiency due to this change in c may be com- 
puted as follows: 

/ / \ k ~ 1 



/ / \ AB 

^ = i-(r+])- (42) 

dr, 3 I I \ab i i ^AB ,.„. 

&;=(r+7)* l0 *(r+7)i? (43) 

= ^ (1 - ,0 log. (~^ 

7^3 C» L C v 7? 3 \1 — Z/ J 

This expression would give the fractional variation of effi- 
ciency — -, due to the fractional change in the value of c v . ( — -) • 
773 \ c v J 

TEMPERATURE AFTER EXPLOSION 

To find the temperature after burning, on the assumption that 
no heat is taken away; the intrinsic energy after burning is 
made equal to the heat per cubic foot plus the intrinsic energy 
before burning. If there is an assumed loss of heat to the jackets 
of 20 per cent, of the heat of combustion this same method is 
used with 80 per cent, of the heating value of the fuel. 

U 2 = U 1 + H X (100 - per cent, loss to jackets) (45) 

The intrinsic energy at any point is given by 



C T bT 2 

U = J c v dt = a'T+ ~y 



(46) 



Hence 

bT 2 b 

a'T 2 + -—- + H(l - loss to jackets) = a'iT s + ~ T s 2 (41) 

The values of a'T 2 + ^T 2 2 and of a\Ts -+- ~^T 3 2 are sometimes 

plotted as shown in Fig. 184 so that for a given temperature T 2 
the heat contained in the gas is read from the figure. After add- 
ing H or a portion of H to this, the curve for the burned mixture 



390 



HEAT ENGINEERING 



will give T 3 corresponding to the heat contained in each pound, 
or cubic foot, the intrinsic energy. 

To find the temperature at 4 equation (37) is used. 

If now the mixture at point 1 is assumed to be at 626° abso- 
lute, the temperature at the various points may be found by 
the methods given above. Assume the gases are those mentioned 
on p. 380 and that the clearance is 25 per cent. Of course T\ 
would be worked out by equation (35) if not known. Using 



100 
rf80 

.9 
i 60 

J-. 

y 

1 20 

M 


















/ 



































































500° F lOOO'F 1500°F 2000°F 2500°F 3000°F 3500F 4000°F 

Abs. Abs. Abs.. Abs. Abs. Abs. Abs. Abs. 
Degrees Absolute 

Fig. 184. — Plotting of intrinsic energy for variable specific heat. 



data for the mixture on p. 383 in equation (37), the following 
results : 

T 
12.95 X 0.0125 X 2.3 log ^ + 12.95 X 0.191 X 10 ~ 5 [T* - 626] 



626 



^ X 55.7 X 2.3 log 5 



log T 2 + g 6.66 X 10" 5 T 2 = 2.797 + 0.0417 + 0.309 = 3.148 
Assume log T 2 = 3.148 
T 2 = 1400 

This is of course too large due to the second term. 

Assume T 2 = 1200 
3.080 + 0.080 = 3.160 

slightly too large Try T 2 = 1150 

3.061 + 0.077 = 3.148 

This is correct. 



INTERNAL COMBUSTION ENGINES 391 

The value of n by equation (38) is 

_ log 626" + log 5 _ 0.265 + 0.699 _ 
71 ~ log 5 ~~ 0.699 - 1 "* 

C 
This is the same value found for the ratio of -—■ which was given 

C v 

as 1.38. That the value of 1150 may be seen to be correct the 

following computation is made by equation (4). 

JW.g)- 1 - 626(f) ' 38 =1150 

To find T 3 the correct method is to equate the intrinsic energy 
at 3 to that at 2 plus the heat of combustion. Since the composi- 
tion of gas after burning is not the same as that before, the 
specific heat is not constant during burning nor are a' and b 
constant during this change. Hence the above method is the 
only one which may be used to find T 3 . 

a' m T 2 + \t<? + H = a'iTt + ~T^ 

12.95 X 0.0125 X 1150 + ^^ X 0.191 X 10~ 5 (1150) 2 + 12.95 

X 77.6 X 0.80 =12.95 X 0.0128 XT 3 + ^^ X 0.306 X10" 5 TV 



7Y + 8360r 3 = 50,700,000 
T 3 + 4180 = 8260 

T z = 4080° F. 

T 4 is found by equation (37) used for the determination of T 2 . 
12.95 X 0.0128 X 2.3 X log ^^ + 12.95 X 0.295 X 10 -5 

X [4080 -r 4 ]=^X 55.7 X 2.3 log 5 

log T 4 + 10.04 X 10~ 5 T, = 3.610 + 0.408 - 0.301 = 3.717 

For log T, = 3.71 
T 4 = 5150 
This, of course, is too large. 

Try T 4 = 3000 
3.478 + 0.302 = 3.780 



392 



HEAT ENGINEERING 



Too large. 
Too small. 

Too large, 
which gives 



log 



n = 



Try 7% = 2500 
3.399 + 0.252 = 3.651 

Try T* = 2750 
3.440 + 0.280 = 3.720 

Try Ti = 2740 
3.438 + 6.275 = 3.713 
1080 

0.172 + 0.699 



2745 



+ log 5 



log 5 



0.699 



1.23 



In the calculation for the value of n on the two lines, it has 
been found that on the compression line n = 1.38 while on the 
expansion line n = 1.23. From actual tests as mentioned on page 
214, n = 1.288 on compression and 1.352 on expansion. The 
value of n is found by plotting the logarithms of the volumes 
and pressures of various points and drawing a straight line 
the slope of which will give the value of n. These temperatures, 
626, 1150, 4080 and 2745, are not much higher than those found 
in practice. 

TEMPERATURE ENTROPY DIAGRAM 

The temperature entropy diagram for a gas-engine cycle may 
be easily constructed from the indicator diagram by the method 



3.00 
















1 


12 












A 

.32.00 






13 










0) 

3 




11 




14 








2 




10 






*»M 






& 1.0 




9 

I 7 






^<^18 


19 


20 






(i 


5 






X. 22 








—^— -i__ 


2 


\23 
\24 




. 



1.0 2.0 3.0 4.0 5.0 6.0 

Volume in Inches 

Fig. 185. — Card prepared for T-S analysis. 

shown below. The mean indicator card, Fig. 185, is constructed 
from a number of cards in the same manner as that used in the 
Hirn's or Temperature Entropy Analysis, and then the lines of 
absolute zero of volume and of pressure are laid off after meas- 



INTERNAL COMBUSTION ENGINES 



393 



uring the clearance and calibrating the indicator spring. A 
series of points is then marked and the distances from the axes 
are measured in inches. These are tabulated as shown: 



Points 


Pres- 
sure in 
inches 


Vol : 
ume in 
inches 


Px 

Pi 


Vx 

Vi 


log^ 
pi 


log v7 


7 1 V* 

fclog — 


S x - Si 


T x 
Ti 


1 


0.15 


6.00 


1.00 


1.000 


0.000 


-0.000 


-0.000 


0.000 


1.000 


2 


0.178 


5.25 


1.18 


0.873 


0.072 


-0.060 


-0.084 


-0.012 


1.030 


3 


0.224 


4.43 


1.49 


0.737 


0.173 


-0.132 


-0.185 


-0.012 


1.098 


4 


0.282 


3.75 


1.88 


0.625 


0.274 


-0.204 


-0.286 


-0.012 


1.175 


5 


0.350 


3.18 


2.33 


0.530 


0.367 


-0.276 


-0.385 


-0.018 


1.235 


6 


0.447 


2.68 


2.98 


0.447 


0.474 


-0.350 


-0.490 


-0.016 


1.335 


7 


0.563 


2.27 


3.76 


0.378 


0.575 


-0.423 


-0.592 


-0.017 


1.422 


8 


0.669 


2.00 


4.45 


0.332 


0.648 


-0.479 


-0.671 


-0.023 


1.478 


9 


1.000 


2.01 


6.68 


0.333 


0.825 


-0.478 


-0.670 


0.155 


2.260 


10 


1.260 


2.015 


8.40 


0.335 


0.924 


-0.476 


-0.668 


0.256 


2.810 


11 


1.78 


2.025 


11.86 


0.337 


1.074 


-0.472 


-0.660 


0.414 


4.000 


12 


2.51 


2.040 


16.79 


0.339 


1.224 


-0.470 


-0.658 


0.566 


5.680 


13 


2.00 


2.470 


13.35 


0.412 


1.125 


-0.385 


-0.539 


0.586 


5.490 


14 


1.58 


2.940 


10.50 


0.488 


1.022 


-0.312 


-0.437 


0.585 


5.130 


15 


1.24 


3.55 


8.25 


0.590 


0.916 


-0.229 


-0.320 


0.596 


4.860 


16 


1.13 


3.87 


7.52 


0.643 


0.876 


-0.192 


-0.268 


0.608 


4.840 


17 


1.00 


4.23 


6.67 


0.705 


0.824 


-0.152 


-0.213 


0.611 


4.770 


18 


0.89 


4.65 


5.93 


0.775 


0.773 


-0.111 


-0.155 


0.618 


4.590 


19 


0.795 


5.13 


5.30 


0.855 


0.724 


-0.068 


-0.095 


0.629 


4.480 


20 


0.710 


5.64 


4.73 


0.940 


0.674 


-0.027 


-0.038 


0.636 


4.450 


21 


0.60 


5.70 


4.00 


0.950 


0.602 


-0.022 


-0.031 


0.571 


3.800 


22 


0.50 


5.77 


3.33 


0.960 


0.523 


-0.018 


-0.025 


0.498 


3.190 


23 


0.40 


5.83 


2.67 


0.971 


0.426 


-0.012 


-0.017 


0.409 


2.590 


24 


0.30 


5.89 


2.00 


0.980 


0.301 


-0.009 


-0.013 


0.288 


1.960 



Now if one of the points, say 1, be taken as the datum point, 
the ratios of the temperature at various points to the tem- 
perature of this datum point and the change of entropy from this 
datum may be figured. Thus considering the point 2: 

T* _ P2V2 
7\ Vl Vi 
T2 /v*\ fV* 



« _ CBi\ (Yj\ 



(48) 



This ratio holds to the point of explosion since the unburned 
mixture after burning is different in volume, due to the change 
on burning, and it is necessary to change the temperature found 
by the above formula in the following proportion: 

T , = T vol. of mixture 

vol. of burned gases 



394 HEAT ENGINEERING 

In practical application the change in volume due to chemical 
action is so slight that the volume ratio is assumed unity. 

If the temperature of the point 1 be assumed on the T-S 
diagram, the height to other points may be found by multiply- 

T 2 
ing by the ratio 7^- If the height to 1 be made 1 in. the ratio 

-t 1 
T 2 
7p- will give the heights to the various points. Of course the 

scale of temperature is not known until one temperature is 
known. As will be shown, much data can be determined even 
if this scale is not known. 

Now 



Now 



dv dp 

p v p 




1 v * 1 i v* 
s 2 — Si = c p log e — + c v log e — 

Vi pi 


(50) 


s 2 — Si C p V 2 . , Pi 

2.3 c, = c 8 logl ^ + logl0 P . 


(51) 


Si S\ S% Si 





2.3 c v const. 



The constant of this expression only changes the scale of 
entropy if the expression 

k logio tT + lo gio ZT ( 52 ) 

v 1 Pi 

is plotted as the entropy. Fig. 186 is obtained by this method 

and the area of the figure which represents the difference between 

the heat added and that taken away is equal to the work done or 

the work shown by the indicator card. If the area scale of the 

indicator card is known in B.t.u. per square inch the area scale of 

the T-S diagram will be inversely proportional to the areas of 

the two figures. 

W 

Scale< s = ^~ X Scale pu (53) 

r ts 



Scale, 



cu. ft. per in. X lbs, per sq.ft. per in. 

778 



F pv = area of p-v diagram 

Fta = area of T-S diagram (54) 

The scale fixes the heat and efficiency of the cycle, although the 
component scales of temperature and entropy are not known. 
As soon, however, as one temperature is determined the scale 



INTERNAL COMBUSTION ENGINES 



395 



of temperature will be known and from it and the area scale 
the entropy scale in B.t.u. per degree can be found. The area 
abed is the heat accounted for by the card. If the curve be is 
carried out to e so that abef is equal to the heat per card the area 
dcef represents the reduction in the heat of combustion due to 



8.00 
















I 










































7.00 












































l\ 














6.00 














/ 




























4 


\ 














5.00 














A 




























/ | 


vt 


n 












4.00 










] 


l/ 




g M 


























r 


L j 














3.00 












/ 


/j 

22 






















10/ 




/23 


















2.00 






9/ 
























8 > 








24 




















1.00 


1 


j 


























n\ 




























a\ \k 










d\ 


h\ \j 















-0.10 0.0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 
Temperature -Entropy Diagram to Special Scale 

Fig. 186. — Temperature-entropy diagram of the gas engine card. 

absorption from the cylinder walls and incomplete combustion. 
The area degh represents the heat absorbed from the walls of 
the cylinder during expansion. The area ibeg represents the 
work. The theoretical form of the cycle should have been 
nbem. abik is equal to heat absorbed by wall during compres- 
sion. 



396 



HEAT ENGINEERING 



LOGARITHMIC DIAGRAM 

In Fig. 187 the logarithmic diagram for the indicator card 
is drawn and from it the values of the exponents n are found. 

S0.50 



0.00 



1.50 



1.00 









11 """** 
10 


14 


Slope of Line = 
15 

^1" .. 


n =1.25 








9 


^_^ Slope 
<3"" 


of Line = 


n = 1.35 


ii20 












4 


3 - 


^1 



0.1 



0.2 0.3 0.4 0.5 0.6 

Logarithm of Volume in Inches 



0.7 



0.8 



Fig. 187. — Logarithmic diagram of gas engine card. 

TEST OF GAS ENGINE 

As a closing problem on gas engines the following data is 
taken from a report on a test of a producer and two-cylinder gas 
engine reported by Prof. H. W. Spangler in the Journal of the 
Franklin Institute, May, 1893. The data is as follows: 

2 cylinders each 14^ in. X 25 in. 

Coal — by weight — 

Moisture 4 . 20 per cent. 

Volatile matter (5.80 per cent. CH 4 , 

0.73 per cent. H 2 ) 6 . 88 per cent. 

Fixed carbon 80.41 per cent. 

Ash 8.51 per cent. 

100.00 

Sulphur . 74 per cent. 

Time of test 525 min. 

Revolutions 84,425 160.76 r.p.m. 

Explosions 81,673 155.44 ex.p.m. 

Mean gas pressure \ ll A& in. = 0.062 lbs. per sq. in. 

Barometer 14.686 lbs. per sq. in. 

Gas temperature 75 . 5° F. 

Room temperature 81 .3° F. 

Jacket temperature outlet 99 . 23° F. 

Jacket temperature inlet 62 . 42° F. 

Exhaust pyrometer 752 . 6° F. 

Brake load 1148.5 lbs. 

Zero brake load 160 . 

Brake arm 3 . 073 ft. 

B.h.p 92.73 

I.h.p Top cylinder 64.36 

Bot. cylinder 66.80 131.16 

Coal 1069.6 lbs. 



INTERNAL COMBUSTION ENGINES 



397 



Fuel gas by volume C0 2 4.02 per cent. 

2 0.26 

CO 25.38 

H 2 4.51 

CH 4 1.79 

N 2 64.04 100.00 

Exhaust gas by volume C0 2 . . . 15.60 per cent. 

2 .... 2.24 

CO.... 0.28 

N..... 81.93 100.00 



Cooling water 664 lb. in 4 min. 

Gas for ignition tube 840 cu. ft. 

Relative humidity of air 80 per cent. 




Fig. 188. — Average card from test of Otto engine. 

The following data will be computed: 

Heat of coal C = 80.41 X 14,544 = 11,690 

H 2 = 0.73 X 61,500 = 449 

CH 4 = 5.80 X 24,000 = 1,390 

13,529 

Total heat supplied 1069.6 X 13,529 = 14,470,600 B.t.u. 

Carbon per pound of gas: 

Relative carbon from C0 2 = 4.02 X 12 = 48.24 

Relative carbon from CO = 25.38 X 12 = 304.56 

Relative carbon from CH 4 = 1.79 X 12 = 21.48 

Total relative carbon 374.3 

Relative weight of gas : 

From C0 2 4.02 X 44 = 176.5 

2 0.26 X 32 = 8.3 

CO 25.38 X 28 = 710.0 

H 2 4.51 X 2 = 9.0 

CH 4 1.79 X 16 = 28.6 

N 2 64.04 X 28 = 1792.0 

100.00 2724.4 



398 HEAT ENGINEERING 

2724.4 
Hence molecular weight of mixture = -. ' = 27.24 

374 3 
Carbon in 1 lb. of gas = 070X4 = °- 1372 

Carbon per pound coal 

From C = 0.8041 X 1 = 0.8041 
12 0.0435 



From CH 4 = 0.0580 X 



16 0.8476 



Pound of gas per pound of coal = n ' 79 = 6.18 lbs. 

359 
Volume of 1 lb. of gas = 07-04 = 13.1. 

Volume of gas per pound of coal = 6.18 X 13.1 = 81.00 cu. ft. 

Heating value of gas per cubic foot : 

High Low 

From CO = 0.2538 X 337 = 85.5 85.5 

From H 2 = 0.0451 X 345 = 15.5 13.1 

From CH 4 = 0.0179 X 1071 = 19.2 17.1 



Heat per cubic foot, B.t.u 120.2 B.t.u. 115.7 B.t.u. 

Heat in gas per pound coal = 81 X 120.2 =9740 B.t.u. 

9740 
Efficiency of producer = .--.. = 0.72 

(neglecting temperature) loOZy 

= 72%. 

525 
Indicated work 131.16 X 2546 X qq- = 2,920,000. 

2920000 
Indicated thermal efficiency of producer and engine = 1 4,4,70^00 = 20.2 % . 

20.2 
Indicated thermal efficiency of engine = tt^ =0.281 = 28.1%. 

525 
Delivered work = 92.73 X 2546 X^ = 2,070,000. 

Delivered thermal efficiency of producer and engine = 1 447^/^ = 14.30%. 

Delivered thermal efficiency of engine = 7 o*oo = ^-199 = 19.9%. 

525 
Heat in cooling water = 664 X ~r [99.23 - 62.42] = 3,220,000. 

3220000 
Percentage of heat in coal removed by jacket water = TT^jK ^nR ~ 22.2%. 

22.2 
Percentage of heat in gas removed by jacket water — no7\ = 30.9%. 

Air required per cubic foot of gas: 



INTERNAL COMBUSTION ENGINES 399 

The products of combustion for complete combustion and no dilution are : 





Per cent, 
vol. 


C0 2 


N 2 


H2O 


Air 


CO-, 


0.0402 
0.0026 
0.2538 
0.0451 
0.0179 
0.6404 


0.0402 








Oo 


0.0098 
0.4300 




0.0124 


CO... 


. 2538 




0.6080 


Ho 


0.0451 
0.0358 


0.1070 


CH 4 


0.0179 


0.1443 
0.6404 


0.1710 


N 2 
















0.3119 


1 . 2545 


0.0809 


0.8736 



The analysis of exhaust gases shows 15.60 per cent, by volume of C0 2 
and 0.23 per cent. CO. Now 1 volume of CO would produce 1 volume of 

23 
C0 2 . Hence if j^-gg X 0.3119 = 0.0045 cu. ft. be left as CO the propor- 
tions will be as follows: 

C0 2 0.3074 

CO 0.0045 

N 2 1.2464 

H 2 0.0809 

Air Required . 8629 

The exhaust gases contain 2.24 per cent, of 2 and 15.60 per cent. CO2 
hence there must be 

~j^ X 0.3074 = 0.044 cu. ft. 
of 2 per cubic foot of gas, or 

' 044 noi «. * • • 

ft-^r = 0.21 cu. it. of air in excess. 

This gives as the total amount of air per cubic foot 0.21 + 0.8629 = 
1.0739. The amount of air theoretically required is 0.8629 cu. ft. and 
hence the per cent, excess is 

0.21 

0^8629 = 24 - 2 per cent - 

The products of combustion without moisture will then be: 

C0 2 0.3074 cu. ft. 

CO 0.0045 cu. ft. 

N 2 1 .2464 cu. ft. 

H 2 0.0809 cu. ft. 

Air 0.2100 c u. ft. 

Total 1.7492 cu. ft. 

These came from 1 cu. ft. of gas and 1.0729 cu. ft. of air. 

The gas passes from the wet scrubber to the engine and is therefore 



400 HEAT ENGINEERING 

saturated. Moisture in gas is sufficient to saturate same at 75.5° F. and will 
exert a pressure of 0.436 lb. and hence for a barometer of 14.686 lbs. and a 
pressure of 0.062 lbs. gauge or 14.748 lbs. absolute, the moisture occupying 
1 cu. ft. at 0.436 lb. would occupy 

j^| = 0.0296 cu. ft. 

at atmospheric pressure plus gas pressure. 

The air at 81.3° F. is 80 per cent, saturated and would exert a pressure of 

0.527 X 0.80 = 0.422 lbs. per sq. in. 

and therefore the moisture per cubic foot of air under atmospheric condi- 
tions would be 

0.422 

14686 = 0.0287 cu. ft. 

Now the air and gas are not under the same conditions of pressure and 
temperature and the 1.0739 cu. ft. of air above computed must be reduced 
to the same pressure and temperature as the atmosphere. The pressure on 
the gas is 

14.748 - 0.436 = 14.312 lbs. 

and the air pressure is 

14.686 - 0.422 = 14.264 lbs. 

The temperatures are 75.5° F. for the gas and 81.3° F. for the air. 
The amount of air if reduced to the conditions of atmosphere will be : 

1-0739 X«i|±i|xi|i| = 1.09 cu.f, 

The mixture entering the engine with 1 cu. ft. of gas and its molecular 
weight are given below: 

Gas 1 .000 cu. ft. 

Air 1 .074 cu. ft. 

R n /gas .030 cu. ft. 

2 \air 0.031 cu. ft. 

Total 2.135 

The products of combustion per cubic foot of gas and the computation 
for the mean molecular weight are given below: 

Volume V X mol. wt. 

CO 0.0045 0.126 

N 2 1.2464 35.000 

Air ..0.2100 6.02 

f 0.0809 ] 

H 2 -j 0.0300 [ 2.67 

( 0.0310 J 

C0 2 0.3074 13.43 

Total 1.9112 57.246 

57 246 
Mean molecular weight = . qiio = 30.1 



INTERNAL COMBUSTION ENGINES 401 

The values of the specific heats for the burned mixture are given below : 

Gas Value of a 

CO 0.0045 X 0.018 = 0.000081 

N 2 1.2464 X 0.018 = 0.022500 

Air 0.2100 X 0.018 = 0.003780 

H 2 0.1429 X 0.0188 = 0.002690 

C0 2 0.3074 X 0.0202 = 0.006280 

0.035331 

Gas Value of b 

CO 0.0045 X 0.185 X 10~ 5 = 0.0083 X 10" 5 

N 2 1.2464 X 0.185 X 10~ 5 = 0.2320 X 10~ 5 

Air 0.2100 X 0.185 X 10~ 5 = 0.0388 X 10~ 6 

H 2 0.1429 X 0.668 X 10" 5 = 0.1950 X 10" 5 

C0 2 0.3074 X 0.807 X 10" 5 = 0.2460 X 10~ 5 



0.7201 X 10- J 



0.035331 nMnn , 
a = -Tsrnz = 001831 

> - °-^iT 2 — = °- 3762 x 10 " 5 

From this the heat in the exhaust gases at 752.6° F. above 75.5° F. is 

Heat = VfC P dt = V[a(T 2 - Ti) + \{TS - TS)} (55) 

= 1.9112[0.01831(752.6 - 75.5) + Q- 3762 ^ 10 ~ 5 (^^ 2 _ ^ 2 )] 

= 1.9112[12.398 - 1.052] 
= 25.70 

Heat in exhaust gases at 75.5° F. due to evaporation of water: 
Volume of H 2 = 0.1429 cu. ft. 



0.143 X 14.7 X 1 44 
X 535.5 



Weight of H 2 = 1544 — - = 0.0066. 



18 
Heat of vaporization = 1049 X 0.0066 = 6.92 B.t.u. 

Total heat in exhaust gases per cu. ft. = 32.62 B.t.u. 
Heat of combustion in carbon monoxide in exhaust gas 

0.0045 X 337 = 1.518 B.t.u. 

Heat Balance per cu. ft. of gas: 

Heat supplied 120.20 

Heat brought in above 75.5° F. by air = 

1.070[81.3 - 75.5] [0.018] 0.11 

Moisture in gas and air = e^ 43 x 2.67 1.15 

Total Heat Supplied 121.46 

26 



402 



HEAT ENGINEERING 



Heat in indicated useful work (28. 1 

per cent.) 33.8B.t.u. 

Heat in jacket water (30.9 per cent.) 36 . 2 

Heat in exhaust gases I fi ' 

Heat in CO 1.52 

Difference 18.96 



COMBUSTION OF FUELS FOR BOILERS 



27.8% 

29.8 

26.8 

1.3 
14.3 



100.0% 



The combustion of other fuels may be considered at this 
point with advantage. This combustion is practically the same 
as that of gases considered earlier. This union of the elements 
with oxygen develops heat. The heat of combustion and the 
air required are found in the table on p. 377 as was done for 
gases. The solid and liquid fuels are composed primarily of 
carbon, hydrogen, oxygen and sulphur., 

The solid fuels are coal, lignite, peat, wood and certain waste 
materials such as bagasse. These differ in their chemical com- 
position. This composition is found by chemical analysis. 
When the analysis is made in a combustion tube giving the 
percentages of the various constituents, the analysis is called 
an ultimate analysis. If, however, the analysis gives only the 
moisture, volatile matter, fixed carbon and ash, it is known as a 
proximate analysis. This analysis is easily made and is the 
one made by engineers to place a coal. The analysis is given 
in percentages of the various constituents as received and on 
a dry basis, eliminating the moisture. This latter is necessary 
to reduce the effect of the variable moisture to zero. The 
results of analyses are given below: 

Analysis on Dry Basis 





Ultimate analysis 


Proximate analysis 


Heat 




C 


H 





N 


S 


Ash 


Fixed 
C 


Vol. 

matter 


Ash 


Value 




88.0 
81.0 
75.0 
65.0 
61.0 
51.0 


2.0 
4.8 
5.0 
4.5 
6.0 
7.6 


2.0 
4.0 
9.0 
20.0 
28.10 
4.0 


0.8 
1.5 
1.5 
1.0 
1.9 
37.0 


0.50 
0.75 
1.5 
1.5 


6.7 

7.95 

8.00 

8.00 

3.0 

0.4 


85 
73 
56 
44 


7.00 
20.00 
36.00 
48.00 


8.00 
7.00 
8.00 
8.00 


13,750 


Semi Bituminous 


14,700 
13,800 




10,700 


Peat 


9,000 


Wood 








9,000 













At times the analysis is put on an ash- and moisture-free 
basis. This is expressed in percentages of the constituents 
with ash and moisture omitted. 



INTERNAL COMBUSTION ENGINES 403 

The constituents unite with the oxygen according to the 
formulae 

C + 2 = co 2 . 
H 2 + HO2 = H 2 0. 

s + o 2 = so 2 . 

From the atomic weights of the elements, the weights of the 
oxygen required per pound of carbon, hydrogen or sulphur 
may be found, together with the products of combustion. Thus 
1 lb. of carbon requires 3 %2 lbs. of oxygen, 1 lb. of hydrogen re- 
quires 1 % or 8 lbs. of oxygen and z %2 lb. of oxygen are required 
per pound sulphur. If these quantities are divided by 0.232 the 
air required per pound of the various substances is known. If 
C, H and S represent the parts of these substances in 1 lb. of 
fuel, the air required is given by: 

Lbs. of air per lb. fuel = 11.5 C + 34.5 H + 4.35 S. 
This is given also as: 

Lbs. of air = 11.5 C + 34.5 (h - ^) + 4.35 S. 

or approximately 

Lbs. of air = 12 C + 36 (# - ^) • 

The value ^ is substracted from H if is the weight of the 

oxygen in the fuel, since it is assumed that this oxygen is united 
with the proper amount of hydrogen in the form of H 2 0. 
The heat of combustion has been given by Dulong. 

Heat per pound = 14,650C+62,100 (h - g) (56) 

The value given by this formula is approximately that given by 
the bomb calorimeter. The results of burning determined by a 
calorimeter are given in the table on p. 402. 

The products of combustion of solid fuels are principally C0 2 , 
N 2 and excess air. If pure carbon is burned, the C0 2 formed is 
just equal in volume to the oxygen consumed, so that the maxi- 
mum amount of C0 2 is 21 per cent, by volume. As there is 
some hydrogen present and as the water occupies twice the vol- 
ume of oxygen consumed, each per cent, of hydrogen reduces the 
percentage C0 2 because, although the water condenses, the nitro- 
gen of the air left from its burning cuts down the possible per- 
centage. As soon as oxygen appears in the burned gases it 
indicates that excess air has been supplied. This is advisable in 



404 



HEAT ENGINEERING 



many cases as there is danger in having some CO when the air- 
is not in excess and the loss due to improper combustion is greater 
than that due to the presence of excess air. This latter represents 
a loss due to a larger quantity of hot gas sent from a boiler or 
gas engine. It is shown that at least a 30 per cent, excess is 
necessary to prevent the formation of CO. The curve of Fig. 
189 illustrates the relation between per cent. C0 2 and total air 
supply as a per cent, of the theoretical air supply. It is found 
that when the per cent. CO2 is between 10 and 15 per cent, the 
best results are obtained. 

In analyzing gases from furnaces a percentage of C0 2 of 15 
per cent, with 3 or 4 per cent, of 2 represents good results. 



35j£ 
30 % 

2Sf 

l ,20fc 

j 

1 

IO56 

5# 









1° 


















IS - 

I 


















1 © 

1 Q> 
1 ° 








































\^§ 


>^l 


















^<Oj, 


^i£e 























0.8 



.4 S 



0.2 



0.0 



50 % 25/» 50 jj 750 100 125 150 175 200 225 250 

Deficiency of Air Excess of Air 

Fig. 189. — Curve of relation between per cent. C0 2 and excess of air. 

PROBLEM 

The analysis of the exhaust gases from a boiler gives the exact excess air. 
Thus for a coal of the following analysis: 

Per Cent, by Weight 

C 80.0 

H 2 5.0 

2 10.0 

S 0.5 

Ash 4.5 

100.0 
and flue gases of the following analysis : 



INTERNAL COMBUSTION ENGINES 405 

Per Cent, by Volume 

C0 2 12.0 

CO 0.5 

2 5.0 

N 2 82.5 

The amount of air may be worked out. The oxygen and carbon in the ex- 
haust gas may be found by remembering that from Avogadro's Law, equal 
volumes at the same temperature and pressure contain equal numbers of 
molecules. Hence each volume may be considered to contain one molecule. 
The weights of carbon and oxygen of the gas are found as follows : 

Carbon from C0 2 =0.12 X 12 = 1.44 
Carbon from CO = 0.005 X 12 = 0.06 
Total carbon 1 . 50 

1.44 

Per cent, carbon burned to C0 2 = = 96 per cent. 

1.50 P 

Per cent, carbon burned to CO = — — = 4 per cent. 

1.50 P 

Oxygen from C0 2 = 0.12 X 32 = 3.84 

Oxygen from CO = 0.005 X 16 = 0.08 

Oxygen from 2 =0.05 X 32 = 1.60 

Total oxygen required 5 . 52 

5.52 
Oxygen per lb. carbon = — — =3.68 
1.50 

1 fin 
Pounds excess of oxygen supplied per pound carbon = — — = 1.067 

1.50 

Per cent, excess of oxygen in total oxygen shown by analysis = 

160 X 100 

— — = 29 per cent. 

552 

The gas analysis does not show the oxygen required for the hydrogen and 
sulphur and these must be added to that for the carbon and from the same 
the oxygen of the coal is subtracted. The oxygen required is given by: 

Oxygen for carbon as burned and excess 

oxygen per pound coal* = . 80 X 3 . 68 = 2 . 944 

Oxygen for hydrogen = . 05 X 8 . = . 400 

Oxgen for sulphur to S0 2 = 0.005 X 1.0 = 0.005 

3.349 
Oxygen in coal . 100 

Net oxygen per pound coal 3 . 249 

3.249 

Air per pound coal = — = 13.95 

0.232 

™ , - . . , . .. , 0.29 X 2.944 

bxcess ot air in total air supplied = - = 26.25 per cent. 

3.249 



406 HEAT ENGINEERING 

Products of Combustion : 

44 

Carbon dioxide 0.8 X 0.96 X — =2.810 

iz 

28 

Carbon monoxide 0.8 X 0.04 X — = 0.075 

iz 

Water vapor 0.05 X 9 = . 450 

S0 2 . . . ! 0.005 X 2 = 0.010 

Excess air 1.067 X 0.80 X = 3 . 670 

Nitrogen (13.95 - 3.67) 0.768 = 7.890 

Ash = 0.045 

Total weight = 14 . 950 

This checks the 1 lb. of coal and the 13.95 lbs. of air. 

The moisture brought in by the air may be treated in the 
same manner as that used in the problem of p. 400. The 
heat carried up the stack may be computed in a manner similar 
to that used in finding the heat in the exhaust gases of a gas 
engine. This quantity when divided by the heat of combustion 
of coal gives the percentage loss in the stack. This amounts 
to about 10 or 15 per cent, when there is just sufficient air, while 
with 100 per cent, excess it amounts to 20 per cent. A deficiency 
of air means a loss due to incomplete combustion. After the 
carbon burns to C0 2 a further passage through a bed of hot coals 
changes the CO2 to CO and the water from the burning of the 
hydrogen may be dissociated. On mixing the CO with hot air 
and having the H 2 and O mix after cooling, in the presence of a 
flame the gases burn and give out their heat of combustion. 
With an insufficient or a cold air supply above the fire the CO 
may not burn. If volatile hydrocarbons are driven off in the 
form of smoke these must be mixed with air in the presence of 
incandescent fuel to prevent smoke formation. 

SURFACE COMBUSTION 

To cause gas to burn completely with the theoretical amount 
of air present or a little above the requisite amount of air the 
method of surface combustion has been suggested. In this a 
mixture of gas and air is delivered into a tube of a burner and dis- 
charged into a receptacle filled with small pieces of refractory 
material. If the mixture is lighted and its velocity through 
the tube is more rapid than the speed of burning, the flame will 



INTERNAL COMBUSTION ENGINES 



407 



not enter the tube and cause an explosion but the mixture will 
quietly burn and heat the refractory material to brightness unless 
the heat is removed by some method. Fig. 190 illustrates one 
form of burner. In the Bone-Schnabel boiler this method is 
applied, the combustion taking place in the boiler flues which are 
cither filled with small pieces of a refractory substance or rods 
of the same. The heat of combustion then heats these substances 
so that the heat is transmitted to the tube surface by radiation 
and by conduction so readily that not only is the capacity in- 
creased but the efficiency is greatly increased due to the low 
temperature of the exhaust. 
The combustion is complete. 
In tests quoted in the Journal 
of the A.S.M.E. for Jan., 
1914, p. 09, G. Neuman re- 
ports a test on one of these 
boilers showing an evapora- 
tion of 16.4 to 30.75 lbs. per 
square foot of surface and an 
efficiency of 90 per cent. 

The porous filling not only 
causes the gas to move at a higher velocity relative to the tube 
but these heated objects maintain combustion and thoroughly 
mix the gas and air preventing stratification. 




Fig. 190. 



-Burner for surface com- 
bustion. 



TOPICS 

Topic 1. — Sketch the cycles of Beau de Rochas, Atkinson, Otto, Lenoir, 
and Diesel. Explain the action on each line and reduce expressions for the 
thermal efficiency of each cycle. 

Topic 2. — From the efficiency of the Otto cycle on the air standard 



= 1 



T t -T b 

Td — T c 



Reduce the expressions 



= 1 



~ = 1 



(ft)- 



1 + 



Give reasons for the impossibility of reducing the expressions for efficiency 
of the other cycles by this method. What is meant by air standard? 

Topic 3. — Explain the action of a four-cycle gas engine and a two-cycle oil 
engine. Draw cards from each. Explain how these engines are governed. 
Explain the methods of ignition. 



408 HEAT ENGINEERING 

Topic 4. — What affects the rapidity of combustion? Draw an indicator 
card showing slow burning. Sketch a curve showing the variation of gas 
temperature in a four-cycle gas engine. Is there much cyclic change in the 
metal of the cylinder walls? 

Topic 5. — Mention the various fuels used in internal combustion engines. 
Give some of the peculiarities of each. What are gas producers ? What are 
the two general types? Sketch one of them. 

Topic 6. — Derive the expressions for finding the amount of oxygen and air 
to burn a cubic foot and a pound of C 2 H 4 together with the expressions for 
the products of combustion. Show how to find the heat per cubic foot if 
the heat per pound is 21,400 B.t.u. Find the specific heats per cubic foot if 
c p = 0.404 and c v = 0.340. On what laws are these based? 

Topic 7. — Derive the formulse 

B - 1544 



Mol. wt. 



mol. wt. 

Z mol. wt. X vol. 
2 vol.~~ 



Topic 8. — Derive the formulse for finding the temperatures at the corner 
of the Otto cycle assuming the value of c v to be a constant. 

Topic 9. — Derive the equation for the adiabatic of a gas when c v = a' + 
bT. Derive the equation for n on this line in terms of the temperatures. 

Topic 10. — Explain the method of finding temperature at the ends of 
compression and expansion. 

Topic 11. — Explain by formulse the method of finding the temperature 
after explosion. 

Topic 12. — Explain the construction of the T-s diagram for the gas engine 
deriving the expressions for the construction of the figure. 

Topic 13. — Explain the construction of the logarithmic diagram of the 
gas-engine card. What data can be found from this card? What data must 
be known to construct the card? 

Topic 14. — Explain the method of finding the amount of dilution from the 
analyses of the fuel gas and exhaust gas. 

Topic 15. — Explain the method of finding the efficiency of a producer. 

Topic 16. — Explain the method of finding the heat loss in the exhaust gases 
and in the jacket water. 

Topic 17. — Explain the method of finding the heating value of a gas from 
it chemical composition. Explain how to find the heat equivalent of the 
work. 

Topic 18. — Explain how to find the amount of air per pound of fuel in a 
boiler test from the coal analysis and gas analysis. What is Dulong's 
formula? What is surface combustion? Explain the construction of the 
Bone-Schnabel boiler. 

PROBLEMS 

Problem 1. — The clearance of an engine is 30 per cent, of its longest stroke. 
The initial temperature is 150° F. The heat per cubic foot of mixture 



INTERNAL COMBUSTION ENGINES 409 

is 70 B.t.u. The radiation loss during explosion is 35 per cent. Find the 
temperatures at the corners of Otto cycle and Atkinson cycle assuming the air 
standard. Find the Carnot efficiency of each, the theoretical efficiency and 
the type efficiency. 

Problem 2. — A Diesel engine has a final pressure of 600 lbs. per square 
inch. What is the clearance to give this from atmospheric pressure at the 
beginning of the stroke (n = 1.38). If the original volume including 
clearance is 1 cu. ft. of air at 160° F., find the weight of crude oil given in 
this chapter which could be burned by this air with 25 per cent, dilution. 
If 75 per cent, of the heat is available find the temperature and volume 
after burning, using the air standard. 

Problem 3. — Find the products of combustion of 1 lb. of crude oil burned 
in a Diesel engine with 15 per cent, dilution. Find the temperature resulting 
therefrom with a 25 per cent, loss to the jacket considering the actual com- 
position of the gas and assuring the pressure constant. 

Problem 4. — Find the temperature Ti if the clearance is 30 per cent, and 
the gas and air outside has a temperature of 60° F. with c v per cubic foot 
0014 and the heat per cubic foot 78 B.t.u. Find the other temperatures 
assuming 20 per cent, loss to jacket. 

Problem 5. — Assume a blast-furnace gas of form given in this chapter 
mixed with 10 per cent, excess air. Find the cubic feet of air per cubic 
foot of gas. Find the value of T 2 considering the actual composition of 
the mixture if Ti = 600° abs. and the clearance is 25 per cent. Find T 3 
with 10 per cent. loss. Find T 4 . 

Problem 6. — Sketch an indicator card from an engine with 25 per cent, 
clearance and draw the T-s diagram and logarithmic diagram. Find the 
heat added if the actual efficiency is 28 per cent. Find the scales of the 
figure if T\ is found to be 125° F. What is the maximum value of T\. 

Problem 7. — The following results are obtained from a test: 

Coal Producer Gas Exhaust Gas 

C 80 . per cent. C0 2 4.0 per cent. Co 2 15.8 per cent- 

CH 2 6.0 per cent. 2 0.5 per cent. 2 2.5 per cent- 

H 2 0.5 per cent. CO 25. 5 per cent. CO 0.2 per cent- 
Ash 10.0 per cent. H 2 5.0 per cent. N 2 81. 5 per cent- 

Moisture 3.5 per cent. CH 4 . . . . 1.8 per cent. 

N 2 63.2 per cent. 

100.0 100.0 100.0 

Time of test 1,200 min. Exhaust temperature. 800° F. 

Revolutions 240,000 Jacket temperature 70° 

Explosions 110,000 Jacket temperature.. . 120° 

Gas pressure 2 in. water Coal 4,000 lbs. 

Barometer 29.9 in. Gas 320,000 cu. ft. 

Gas temperature. . . 80° F. I.h.p 200 

Air temperature ... 70° F. B.h.p 160 

Relative humidity . 60 per cent. Cooling water 178,000 lbs. 

Compute the various losses and efficiencies and make a heat balance. 
Problem 8. — The heating value of 1 lb. of coal as fired is 14,680 B.t.u. 



410 HEAT ENGINEERING 

The equivalent evaporation from and at 212° F. per pound of coal is 11.10 
lbs. Find the efficiency of the boiler. 

Problem 9. — In Problem 8 the following was found: The flue gas analysis 
by volume gave 0.18 per cent. CO, 6.49 per cent. C0 2 , 13.09 per cent. 2 , 
and the coal analysis by weight was as follows : 

Moisture 1.0 per cent. 

Ash 6.9 

Hydrogen 2.8 

Oxygen 4.2 

Carbon 83 . 

Sulphur 0.8 

Nitrogen 1.3 

100.0 

Find the amount of air per pound of coal. Find the excess air. Find the 
heat value by Dulong's formula and check with value by calorimeter given 
in Problem 8. 

Problem 10. — In Problem 9 find the composition of the exhaust gases per 
pound of coal assuming that the air at a temperature of 75° F. has a wet 
bulb temperature of 67° F. 

Problem 11. — With the results of Problem 10 find the heat carried away in 
the flue gases due to a temperature of 427° F. and the heat available from the 
CO present. Express these as percentages of 14,680 B.t.u. 

Problem 12. — With coal at $4.40 per ton what is the cost of producing 
1000 lbs. of equivalent evaporation, using data of Problem 8. Equivalent 
evaporation from and at 212° F. is equal to the actual evaporation multiplied 
by the factor of evaporation. Factor of evaporation is the amount of steam 
evaporated from and at 212° F. by the same amount of heat as will evaporate 
1 lb. of water at the given feed temperature into steam at the given 
condition for the boiler. This is given 

, _ q'i -h-xiri - g' = ii — q'o 
r 2 i2 r 2i 2 

In this problem find the factor of evaporation for a gauge steam pressure 
of 121.3 with the barometer of 29 in. if the feed temperature is 105° F. and the 
quality of the steam is 65° F. superheat. Find the cost of producing 1000 lbs. 
of actual steam. 



CHAPTER X 
REFRIGERATION 

AIR MACHINES 

The refrigerating machine operates by placing a substance 
in such a condition that its temperature is above the tempera- 
ture of a water supply, and after the removal of heat from the 
substance by this supply the condition of the substance 
is so changed that it will abstract heat from a body of 




Fig. 191. — Air refrigerating machine. 



low temperature. The temperature of the substance is changed 
by changing the pressure on the substance. This is accomplished 
by machines in which a gas such as air or a vapor such as ammonia 
or carbon dioxide is used. The air machine is shown in Fig. 
191. Air is compressed in the cylinder A from a pressure p\ 
to a pressure p 2 . At the pressure p 2 it is discharged through a 

411 



412 



HEAT ENGINEERING 



Compressor 



pipe system B which is cooled by water from the supply C. In 
this case the cooling of the air reduces its volume so that when this 
air is taken to a second cylinder D it occupies less space. The air 
is then allowed to expand in the cylinder D to the original pres- 
sure, doing work at expense of its intrinsic energy, and hence its 
temperature falls so much that on entering the coil E in the tank 
or room F, this air will remove heat from the brine in the tank or 
from the air, if placed in a room. This air is taken back to the 
compressor after removing heat from the room F. The cycle is 
shown in Fig. 192. At times the air is discharged into the room 
F instead of passing through the coil and air is sucked from the 

room. The system is then called an 
open system to distinguish it from 
that of Fig. 191 which is called a 
closed system. 

If the temperature of the air 
sucked from the cool room F is T\ 
and the pressure is pi, the condition 
of 1 lb. of air is shown by the point 1 ; 
the volume 4-1 is sucked in on the 
line 4-1 if clearance is neglected. 
This air is compressed along the line 
1-2 and the line is practically an 
adiabatic. The temperature T 2 is 
given by 

This air occupies the volume 3-2 
and the card 4123 represents the 
work done on the compressor A. 
The air is forced out and is cooled 
off in the cooler B to a temperature 
T h fixed by the temperature of the 
water at the point of discharge of the air. In a countercurrent 
cooler this may be less than the temperature of the outlet 
water. It is about 10° F. above the temperature of the water 
at the point. When this air passes into the expansion cylinder 
its volume will be shown by the point 5. The admission line 
is 3-5. The air expands to the lower pressure and its condi- 
tion is shown by point 6. This action is also adiabatic, so that 




Fig 



Combined Cord 

192.— Cycle of the air 
machine. 



'.-r.©^ (2) 



REFRIGERATION 413 

Th T 2 
or yr = y \*) 

Te=T b ^ (4) 

It is thus seen that Ti is fixed by the temperature of the refrig- 
erator, being the room temperature in the case of an open system 
or 10° to 15° F. less for a closed system; while T 5 is fixed by the 
temperature of the cooling water, T 2 and T 6 are fixed by the pres- 
sure ratios after 7\ and T 5 are found. 

The work done on the compressor is 4123 while that done by 
the air in the motor is 3564. The difference or 1256 shown by 
combining the cards is the net work required. This work is 
also the difference between the heat on the top and bottom lines 
since there is no heat on the curved lines. 

AW = c p {[T 2 - T b ] -[T!- T e ]} (5) 

It is seen that heat removed by the cooler is equal to the heat 
removed from the refrigerator plus the work done on the 
substance. 

But the heat to change the volume from 2 to 5 is the heat re- 
moved by the cooling water while that added from 6 to 1 is 
that taken from the refrigerator. Hence 

Qref. = c p [T 1 - ffi] (6) 

and Q cooler = c p [T 2 - T 6 ] (7) 

The efficiency or better the refrigerative performance is the 
amount of refrigeration per unit of work. This is given by 

CAT, - U (8) 



But 

since 
and 



C P {[T2- 
1 


rj- 


[Ti- 


Ts]} 


T 2 - 
Tt- 


T 5 
■T 6 


- 1 




T 2 


-T b 

- T 6 

T, 
T 5 
T 2 


T 2 
" T, 

T 2 
~ Tt 

T 6 
~ T x 


' T 6 





414 HEAT ENGINEERING 



Hence 1 1 

or 



Ti _ i Ti _ ■ i (») 

~ m /T7 Or ^ fj, v-L^J 

These expressions are equal to the lower temperature on one 
of the adiabatics divided by the difference in temperature on 
that adiabatic. The expression is greater than unity. 

The refrigeration produced by any machine is usually measured 
in tons of refrigeration per 24 hrs. The amount of heat liberated 
when 1 lb. of ice melts is 144 B.t.u. (best value 143.5 B.t.u.), 
and hence the heat equivalent to 1 ton of refrigeration is 288,000 
B.t.u. in 24 hrs. which is 200 B.t.u. per minute. The number of 
pounds of air required for this refrigeration is 

Lbs. of air per min. per ton in 24 hrs. = — r^ =^y = M a (11) 

Cp[l l — 1 6J 

The amount of cooling by the air cooler is 

Q c = M a c p [T 2 - T 6 ] (12) 

The water required for this is given by 

q o q i 

M w = weight of cooling water per min. per ton in 24 hrs. 
q' = heat of liquid of cooling water at outlet temperature t 
q'i = heat of liquid of cooling water at inlet temperature t { . 

The horse -power required for the production of 1 ton per 24 
hrs. is 

778 
h.p. = M a c p [T 2 -T b - (TV- T*)] 33^ (14) 

ft x 778 
~ Vr X 33000 { J 

778 
= W- " <2 J 33000 (16) 

The volume of the compressor is Vi while that for the ex- 
pansion cylinder is V G . These are each given by 

Vi = or — .. , . — — (17) 

Vi Vi X cl. factor 

T . MgBTs M a BT 6 . 

V 6 = — or — w i j + ( 18 ) 

LPe Pe X cl. factor 



REFRIGERATION 415 

The above equations are computed for cylinders without 
clearance and friction, for non-conducting cylinder walls and 
for dry air. These three things affect the results of computa- 
tions, but because their effects are seen by comparison with 
perfect conditions the equations for perfect conditions are given 
although not used in practice. 

PROBLEM 

Assume that a 3-ton machine was desired to operate between 14.7 lbs. 
abs. and 58.8 lbs. abs. and that the temperatures of the cooling water were 
60° F. to 65° F. and that for the cool room was 25° F. It will be assumed 
further that the air from the expansion cylinder is discharged into the room 
to be refrigerated. From the above the following is true: 

Tx = 25° F. = 485° abs. 

T b = 50° + 15° = 65° F. = 525° abs. 

(assuming a counter-current flow) 

/58 8\ — 
T 2 = 485 Ij^f) 1A = 485 X 1.486 = 720° abs. = 260° F. 

485 
Tt = 525 X.720 = 354° abs. = - 106° F. 

485 

= 2.063 



,/r 720 - 485 
Q r = 200 X 3 = 600 B.t.u. per min. 

^° = O 24r485 — 3541 = 19-08 lbs. of air per min. 

Qcooier = 19.08 X 0.24[720 - 525] = 892 B.t.u. per min. 

778 
H.p. = (892 - 600) 33QQQ = 6.86 

XT 600 778 

also H.p. = 2^3 X 33^ = 6.85 

The water required is 

892 
M w = 00 1 _ 10 1 = 59.4 lbs. per min. 

Displacements per minute are: 

19.08 X 53.35 X 485 QQQ 

Vcomp- = 14 7 x 144 = cu ' per mm * 

_ 19.08 X 53. 3 5X35 4 , + 

XJ " = 14 7 X 144 = CU ' per mm * 

The machine considered in the problem above is one in which 
the discharge has been into an open space. Such a system may 
be called an open system. If now the discharge takes place 
in a closed pipe line (the closed system) it would be possible 
to have the pressure pi higher than atmospheric pressure. Sup- 
pose that pi is made 58.8 lbs. abs. and p 2 , 235.2 lbs. abs. so 



416 



HEAT ENGINEERING 



that the ratio — is the same as before. If under these condi- 

tions the above problem is computed the results will be the same 
except the displacements which will be one-fourth of their former 
values. This is due to the higher pressures. This "dense air 
machine," as it is called, is used because it decreases the dis- 
placement of the cylinders and hence the size or speed of the 
machine. This is the only reason for its use. 

The problems must now be investigated for clearance and 
friction. Since friction increases the work of the compressor 
and decreases the work obtained from the expansion cylinder, 
the work done by each of these machines must be considered 
separately, instead of obtaining the net work 1256 as before. 

The work of the compressor 1234 is the same with or without 
clearance. It is equal to 



^--^i^.fi-©^] 



(19) 



n 



jlpiV! - p 2 V 2 ] = jj^j MBIT, - T 2 l 



If n = k this becomes 



W comp = MJc v [T x - T 2 ] 



(20) 



To make this work as small as possible, jackets may be used 
on the cylinder changing k to n and making n as low as 1.38. 



5' 5 2 2' 


XX-.. 


6' 6 1 



Fig. 193. — Effect of friction in air machines. 

If the friction fraction is /, the work must be multiplied by 
(l+/)or 



n 



1'2'34 = work of compression = (1 + /) ^^MBIT, - T 2 ] (21) 

For the expansion cylinder the work obtained is (1 — /') times 
the work without friction. This work is independent of the 



REFRIGERA TION 417 

clearance if the expansion and compression are each complete. 
The work then becomes : 

35'6'4 = work of expansion = (1 - f)p 6 V 6 ^b[[l- (-)"** 1 

or work = (1 - f')MJc p [T 6 - T 6 ] (22) 

The value of n in the expansion cylinder is k as it is desired 
to have as low a value of T & as possible and this is accomplished 
by having as large a value of n as possible. By lagging the 
expansion cylinder and omitting any jacket this n is made k 
as no heat is taken away. 

The difference of the two expressions for work will give the 
work required from the outside to drive the machine. 

Of course with n on compression (1.38) different from that on 
expansion (1.4) the equality of cross products 

T,T, = T 2 T 6 

is not true. T 2 and T 6 must each be found. 

T 2 = T^f^ (23) 



- ">© 



k-1 



T~ (24) 



The effect of clearance air on the temperatures is important 
to consider. The air left in the cylinder at the end of compres- 
sion is at a temperature T 2 . When this expands down to atmos- 
pheric pressure the temperature is T\ so that the fresh air enter- 
ing at this temperature mixes with it. In the expander the 
air in the clearance space at temperature T 6 is compressed to 
the temperature T$ by complete compression and so mixes with 
air from the cooler at that temperature. If this compression 
were not complete this temperature T$ would not be reached 
and the effect of this would be to give a different temperature 
of the mixture at the beginning of expansion. This incomplete 
compression would require an excess amount of air from the 
supply to fill the clearance space. This free expansion into the 
clearance space would warm the air in the cylinder and produce 
a higher temperature at the beginning and at the end of expan- 
sion. It would mean a decrease in refrigerating effect and would 
also decrease the work obtained from the expander per pound of 
air used. 

27 



418 HEAT ENGINEERING 

If the expansion were incomplete the temperature at the end 
of expansion would not be as low as T 6 due to the pressure range 

— being less- than — or — and as a result the temperature at 
Pe' Ve Vi 

the end of free expansion or the discharge temperature would be 
higher than TV This reduces the refrigerating effect. The 
work obtained from a given amount of air has been decreased 
by the incomplete expansion and compression. For these two 
reasons the performance is decreased by a very material amount 
and the endeavor is made to have complete expansion and 
compression. 

To eliminate the effect of clearance the compression in the 
expansion cylinder is such that the pressure at the end of com- 
pression is equal to pressure of the compressor and the expansion 
is carried down to the back pressure. The action is then equiva- 
lent to a cylinder without clearance. 

The moisture in the air has an effect on the efficiency of air 
refrigerating machines. The compression of moist air may cause 
some of the moisture to condense giving out heat and warming 
the air while a further condensation and even freezing in the ex- 
pansion cylinder causes the liberation of additional heat reducing 
the refrigerating effect. Air will be removed in short time. 

Suppose that air of relative humidity p, temperature 7\ and 
pressure pi occupies the volume V\. The weight of the air is 

pa = sat. steam pressure corresponding to t\. 
The weight of the moisture M m is 

M m = pwiFi (26) 

mi = weight of 1 cu. ft. of steam at temperature t x . 
This may also be approximated by the formula 

M m = fp (27) 

Dm __ 1544 _ 1544 _ g5 g 

m mo!, wt. steam 18 

For the complete mixture of air and moisture 

(M a + M m )B mx T=pV (28) 



REFRIGERATION 419 

After compression to the pressure p 2 the volume becomes 

F.-FxfgT (29) 

and T 2 = !ri(^)~*~ (30) 

Now -W 5 = weight of moisture in 1 cu. ft. (31) 

V 2 

hence, if 

ra 2 = weight of 1 cu. ft. of saturated steam 

Mrn 

— - = P2 = relative humidity after compression (32) 

p 2 is usually less than unity because of the high temperature 
at the end of adiabatic compression. 

The work of compression of the mixture of air and vapor is 
considered to be that of a perfect gas as the steam in the air is 
superheated. If adiabatic this becomes: 

= j lM a c a + M m c m ](T 2 - T x ) (33) 

c a = specific heat at constant pressure for air. 

c m = specific heat at constant pressure for moisture. 

1 1544 O _ 
Cm - 778 X lg X a3 - 0.48. 

If this air is cooled in the cooling coil to a temperature T 5 , 
the moisture may or may not be sufficient to saturate the air. 
To find out the amount, it is known that 

Vs - V 2 J (34) 

Mrn 

and Vs /or\ 

— - P3 (35) 

M 
If m 3 < ~ (36) 

n 

M m — m 3 V f 3 = amount condensed. 

If the moisture condensed is removed by a separator, the 
weight of the material left is 



420 HEAT ENGINEERING 

M a + m 3 V 3 (37) 

in place of M a + M m 

The volume of this is 7' 8 = MaBaT \ (38) 

The value of Q c is 

Qc = M a c p [T 2 - T 3 ] + M mCpm [T 2 - T 3 ] + [M m - m 3 V' 3 ]r 3 (39) 

The moisture which enters the expansion cylinder is practi- 
cally all condensed and frozen by the cool walls of the cylinder 
liberating heat represented by C. 

C = m 3 V' 3 [c p (T 2 - T 32 ) + r 32 + 144] (40) 

This is gradually restored during expansion. 

The volume of the air remaining is 

7" 3 = MaB * T * (41) 

The air and ice in the cylinder now expand with the gradual 
return of the heat C from that produced by the ice formation 
giving: 

M a c va dt + (m 3 V f 3 )c i dt + Apdv = + dC (42) 

M a c va + m 3 V' 3 Cj dt dV dC 

C 

assume dC = — ^ ^- dt 

1 3 — 1 4 

M a c va + m 3 V' 3 Ci . T 3 , , , A , V 3 C . T 3 
g log, ^ + ^ log, ^ = - B{Ts _ T j log e yr 

7 3 __ /P4\ ^3 

c 



i_3 _ /P*\ £_? 

[M a c va + m 3 V' 3 Ci + jr^TY 4 + M a AB] log e ^ 



= M a AB log e - (43) 
Pi 

NOW C va + AB = Cp 

Hence 
(M a c p + m 3 V' 3Ci + Ts °_ T ) log e Y 4 = M « AB lo & f A ( 44 ) 

This is solved for T \ by trial, knowing all the other terms. If 
T 3 — T 4 is assumed from an approximate value of T 4 from 

T *- T *&^ (45) 

the equation leads directly to T± 



REFRIGERATION 421 

The value of 7% then gives the heat taken from the refrigerator 
Q r = M a c p (T 4 - 7\) (46) 

The moisture is not considered at this point as it has been 
condensed and frozen in the expansion cylinder and does not 
enter the refrigerator. 

The work done in the expansion cylinder is best found by the 
following: 



V3 



W e = psVs -{ | pdv - PiVt (47) 

From (42) : 

r , M^+mV'rt+r^TJ C Ti dt ( 

= j (M a c va + m 3 V' 3 c i + y 3 °_ y J [T 3 - T A ] 

Since the moisture has been eliminated on the expansion 
curve p 3 V 3 - p.V, = M a B [T 3 - T 4 ] (49) 

and B +J=A (50) 

•'• w% = 3 ( MaCp + mzYzCi + T 3 - t) [t * " T < ] (51) 

The work required from the outside is 

{1 + f)W. - (1 - f)W. (52) 

and the refrigerating effect is 

Q r = M a c p (T 1 - T A ) (53) 

To apply the above formulae assume that the air enters the compression 
cylinder at 25° F., with a relative humidity of 80 per cent, and is compressed 
to 4 atmospheres. The water in the cooler is sufficient to cool this to 65° 
F. Assume that Vi = 1 cu. ft. 
p 25 ° = 0.065 
P p = 0.065 X 0.8 = 0.052 

[14.7 - 0.052] X 144 X 1 . nQ _ „ 

Ma = 53.34 X 485 = 0082 lbs ' 

M w = 0.8 X 0.00023 = 0.00018; call this 0.0002 

This is a check. 
1 
V* = 1 04)^4= 0.372 

T 2 = 485 X (4)1-4 = 720° abs. = 260° F. 

0.000 2 

0.372 
p > = 0085" = a0 ° 63 - 



422 HEAT ENGINEERING 

or there is not enough moisture to saturate the air. The air is now cooled 
to 65° F. as in the problem on page 415. 

Vz = 0.372 X 720 = 0.272 • 
0.0002 
0.272 0.000753 Q m 



H * 0.000979 0.000979 

Hence none of the moisture is condensed in the cooling coil. If pz were 
greater than unity there would be some condensation. After the determina- 
tion of the amount of condensation a recalculation of Vz would be necessary 
as some of the water has been removed from the mixture leaving the air 
saturated. 

The moisture having a specific weight of 0.00073 would be saturated at 
56° F. so that this moisture is superheated 9° F. 

Approximate value of 7%: 

k-l 0A 

T, = Tz(-) k = 525 X (H) 1A = 354° abs. = - 106° F. 
C = 0.0002[(65 - 32) y 2 + 1071.7 + 144] = 0.2462 
V.. 0.082 [l^ffy = 0.2715 

(0 2462 \ 1 TO 082 

0.082 X 0.24 + 0.0002 X 0.5 + ^p^) ^ log e ^ = w log, 4 

0.02119, Tz 0.082, 
"5^34" lo ^Y, = "778 l0ge4 

log y = 0.161 = log 1.45 

T 4 = ^ = 361° abs. = - 99° F. 
1.45 



We = 778[0.02119][65 - ( - 99)] = 2695 ft.-lbs. 

QA 

Wc = 51 X 14,7 X 144 X 1 Tl - (4) 14 1 = - 3600 ft.-lbs. 

Net work without friction = 3600 - 2695 = 905 ft.-lbs. 

Net work with friction = 3600 X 1.10 - 2695 X 0.9 = 1534 ft.-lbs. 

Q c = [0.082 X 0.24 + 0.0002 X 0.39][260° - 65°] = 3.86 

Q r = [0.082 X 0.24][25 - ( - 99)] = 2.44 

2.44 
Coefficient of performance without friction = qT^t = 2.10. 

778" 
2.44 
Coefficient of performance with friction = Trofi = 1-24. 

778 
2.44 
B.t.u. of refrigeration per ft. -lb. of work = ^ron = 0.00159. 

B.t.u. per h.p.-hr. = 0.00159 X 33,000 X 60 = 3120. 

3120 X 24 
Tons of ice melting capacity per 24 hrs. per h.p. = onOO X 144 = ^-26. 

Assuming 3 lbs. of coal per horse-power hour: 



REFRIGERATION 423 

Tons of ice melting capacity per 24 hrs. per lb. coal = o 94 = 0.00361. 

2.44 
Tons of ice melting capacity per cu. ft. = .. .* oooo = 0.0000085. 

Gallons of cooling water per cu. ft. of piston displacement for a 15° rise 

3 86 
in water temperature = .. - v o qg = 0.0305. 

Gallons per ton of ice melting capacity = n 0000085 = ^590. 

Relative volume of expansion cylinder and compression cylinder = 

Ta 364 

7=r = X5V = 0.745. (Air removed by freezing in short time) 

1 i 4o0 

If this is worked out without moisture the following results 
are found: 

"• - '^X 4 ^ 1 = ° 082 

7\ = 485° abs. = 25° F. 

0A 

T 2 = 485(4) 1A = 720° abs. = 260° F. 
T 3 = 65 + 460 = 525° abs. 

485 
T* = 525 X ^20 = 354 ° abs - = ~ 106 ° F * 

W c = 3600 

354 
We = 3600 X t^h = 2630 

Net work = 1.10 X 3600 - 0.9 X 2630 = 1593 ft.-lbs. 

Qc = 0.082 X 0.24[260° - 65°] = 3.84 

Q r = 0.082 X 0.24[25° - ( - 106°)] = 2.58 

2 58 
Coefficient of performance = ttqo = 1.26. 

778" 

2 58 
B.t.u. per ft.-lb. of work = ^93= 0.00162. 

B.t.u. of refrigeration per h.p.-hr. = 0.00162 X 33,000 X 60 = 3180. 

3180 X 24 
Tons of ice melting capacity per 24 hrs. per h.p. = 9000 v 1 44 = 0-264. 

Assume 3 lbs. of coal per h.p.-hr.: 

Tons of ice melting capacity per 24 hrs. per lb. coal = » v 04 = °. 00368. 

Tons of ice melting capacity per cu. ft. of compression cylinder without 
2 58 
clearance = 144 x 2 000 = ° 000009 - 

Gallons of cooling water per cu. ft. of piston displacement for a 15° rise 
3 84 
in water temperature = 1 r v o~oc = 0.0306. 

Gallons per ton of ice melting capacity = n^ooOOOQ = 34 0()' 
Relative volume of expansion cylinder and compression cylinder = 
S =1 = 0.732. 



424 



HEAT ENGINEERING 



The actual volumes displaced by the pistons will be increased 
by clearance. The volume computed divided by the clearance 
factor will give the actual volume. 

The use of air, although cheap, has the great objection that 
the volume of the compressor and expander are large for a 
given amount of refrigeration. To reduce the size of the 
machines, ammonia or other volatile liquids are used as the 
media. The substances used are NH 3 , S0 2 and C0 2 and certain 
mixtures such as Pictet fluid composed of 97 per cent, of SO 2 
and 3 per cent, of C0 2 . The advantages and disadvantages of 
the various liquids will be discussed later. 

VAPOR REFRIGERATING MACHINES 

In machines using these volatile liquids the heat is abstracted 
not by an increase in temperature of the refrigerating medium 
but by the evaporation of a liquid. The underlying principle 

■ n ,- ,■,,,■,;-,-,• ,-,-,-,-^ 1 ;,-,-,-,-i-,-, - of these machines is to bring 
the vapor to such a pressure 
that the temperature of lique- 
faction is slightly above the tem- 
perature of a water supply, and 
by means of this water the ab- 





Fig. 194. — Arrangement of the 
vapor refrigerating machinery. 



Fig. 195. — P-V diagram of the 
vapor refrigerating machine. Com- 
pressor and expansive cylinder. 



straction of heat is possible. This abstraction condenses the 
vapor and after the pressure on the liquid is so reduced that 
the temperature of vaporization is very low, the liquid may 
evaporate by the heat abstracted from a space at a low 
temperature. The temperature of cooling water fixes the upper 
pressure and the temperature of the refrigerated space fixes 
the necessary low pressure. It is merely a matter of pressure 
regulation to fix the temperature limits. 



REFRIGERATION 



425 




The apparatus used to accomplish this result is shown in Fig. 
194. Fig. 195 represents the PV diagram for the cycle. 

In the compressor A the ammonia or other vapor is com- 
pressed adiabatically from 1 to 2. This will superheat the am- 
monia vapor if dry vapor is taken into the cylinder, while if there 
is considerable liquid mixed with the vapor the compression will 
reduce the amount of this liquid. The first is called dry com- 
pression and the second wet compression. These two lines are 
shown in Fig. 196 which is the T-S diagram of the cycle. 
The line 1-2 in either case is an adiabatic. It is seen more 
clearly from Fig. 196 that 
the temperature is increased 
as the compression takes 
place so that when the com- 
pressed vapor is delivered 
from the compressor into the 
condensing coil or condenser, 
B, this water may abstract 
heat and condense the vapor. 
This occurs from 2 to 3 in Fig. 
196 while in Fig. 195 the line 
2-3' represents the discharge 
from the compressor and 3'-3 
represents the volume of the liquid. At times the liquid is cooled 
to a lower temperature by passing it through a coil cooled by the 
coldest water. This is shown in Fig. 196 by 3-3". 

If now the liquid be allowed to expand from 3 to 4 or 3" to 4" 
on an adiabatic in an expansion cylinder the work 5345 would be 
obtained. This amount would be very slight and the return 
would not pay for the complication. Hence the pressure is 
reduced from p 2 to pi by means of the expansion valve C. This 
throttling action results in a value of the heat content below the 
valve equal to that above the valve. 

U = tV (54) 

or u" = &V" 

This, of course, reduces the refrigerating effect as 

iY > U 

and ^Y" > ii" 

but the elimination of the expansion cylinder has made the appa- 
ratus simpler. The exact loss due to this may be found later. 



Fig. 196 T-S diagram of the cycle 
of the vapor-refrigerating machine. 



426 HEAT ENGINEERING 

After the pressure is reduced the temperature of vaporization 
is so low that heat will be abstracted from the surrounding sub- 
stances, even though they be at a low temperature, and the liquid 
vaporizes. This is accomplished in the refrigerating coil D. 
Here the evaporation of the liquid returns either dry vapor or 
wet vapor to the compressor A and the cycle is repeated. 

The expressions for the heat on the various lines are now 
computed. 

Heat on 1-2 = 

Q c = heat on 2-3 at constant pressure = i 2 — q'z (55) 

Q c = heat on 2-3" at constant pressure = i% — q'z" (56) 

Heat on 3-4, 3-4', 3"-4" or 3"-4'" = 

Q r = heat on 4-1 = ii — u or i x — u f (57) 

But is = ^Y 

.'.Q r = i\ — is or ii — i z " (58) 

Now i\ is the expression for the heat content at either Id or 
lw and i 2 is the heat content at 2w or 2d. It and the other quan- 
tities may be found in ammonia tables. 



i = q' + xr 
or i = q' + r + 



JTsup. 
Cpdt 
Tsat. 



The expression for the heat added on a constant pressure line 
from a to b is 

ib — i a (59) 

as is given under Q r and Q c . 

The equation of the line 1-2 is the adiabatic 

s = constant 

8 = S i + -jjr II we ^ 
I 1 

or pv k = constant if dry. 

k = 1.33 for superheated ammonia. Of course if tables for 
superheated vapor are known, the formula 

Si = s 2 

may be used for the superheated region also. This is really the 
better equation even in the superheated region. 
The work required is the algebraic sum of the heats 



REFRIGERATION 



427 



AW = Q r - Qc = - [Qc - Qr] 
- AW = [i 2 - iA - Hi - U'\ 

= i 2 - ii (60) 

With friction - AW = (1 + /)(i 2 - (61) 

The amounts above are all for 1 lb. of ammonia and to find the 
amount of liquid, per minute, hour or day, the amount of re- 
frigeration in that time is divided by the refrigeration per pound. 
Thus for T tons of refrigeration per day the weight of volatile 
liquid per minute is 

20077 

M = (62) 

^l - iv v J 

The horse-power required is 



h.p. 



M(l+f)(i2-ii) 



42.42 

The weight of cooling water per minute is 

M(i 2 - tV) 



(63) 



M m = 



where 



q - qt 

q' = heat of liquid for condensing water at outlet 
q\ = heat of liquid for condensing water at inlet. 

The displacement per minute of the compressor is given by 



(64) 



D v = 



Mv' 



Mv 



or 



clearance factor 

The displacement of the cylinder is 
found by computing the clearance 
factor as in the case of air com- 
pressors. The lines of expansion 
and compression are of the same 
form. This is shown by Fig. 197, 
the actual form of the card. If V 
for the ammonia on the lower line is 
known 

V 



clearance factor 



(65) 




Fig. 197. — Actual card from 
a compressor. 



D = 



(66) 



i+i-i(m 

The refrigerative effect is given by 

N = ft - ** 

In many cases where rooms have to be cooled, the ammonia 



(67) 



428 HEAT ENGINEERING 

or air is not sent through coils in the room but heat is removed 
from the rooms by circulating cold brine through pipes. The 
brine is cooled by the air or ammonia which passes through a 
coil in a tank through which the brine is pumped. The brine 
is usually a solution of calcium chloride or sodium chloride. 
The advantage of the brine system over the direct expansion 
system is the fact that the break of a pipe would not discharge 
the ammonia into the room and spoil the contents and also the 
fact that the compressor may be shut down for some time and 
the cold brine stored in the brine tank may be used to keep the 
room cool. 

These formulae may be used for any volatile liquid. Even 
water has been used although exceeding low pressures must be 
used to obtain temperatures below 32° F. A problem will be 
computed using NH 3 , S0 2 and CO2 for the media assuming water 
at 50° to 65° F. and a refrigerating room held at 25° F. 

In this problem it will be assumed that 15° is the necessary 
difference of temperature for heat transfer and also that wet 
and dry compression will be tried for the ammonia. For dry 
compression x\ = 1 while for wet compression x 2 = 1. It will 
be assumed that " after cooling" reduces T 3 " to 50° + 15° = 65° 
F. In all cases the volume drawn into the compressor with 
1 per cent, clearance will be assumed to be 1 cu. ft. 

In these problems 



T A = T 7 ! - 25° - 15° + 460° = 470° 


abs. 


T 2d = 65° + 15° + 460° = 540° 


abs. 


T v = 50° + 15° + 460° = 525° 


abs. 


Ammonia: 




Case I. — Wet compression. 




Px = 38.02 lbs. 




p 2 = 153.90 lbs. 




X2 = 1 




Xl = \ s 2 + ~Ti ~ s 7 T Y x 




_ 0.1025 +0.9328 + 0.0483 




1.2017 




i% = 557 




ix = - 23.2 + 0.902 X 564.4 = 485.0 




iz> = 36.5 




M = 6^4 = °- 151 




Q c = 0.151 [564.4 - 36.5] = 78.6 B.t.u. 




Q r = 0.151[485 - 36.5] = 67.8 B.t.u. 





AW = 78.6 - 67.8 = 10.8 B.t.u. 
or = 0.151[564.4 - 485] = 10.8 B.t.u. 



REFRIGERATION 429 

AW with friction = 10.8 X 1.20 = 12.96 

67 8 
Refrigerating effect without friction = YFTo = 6.23 

67 8 
Refrigerating effect with friction = yo~qa = 5.22 

B.t.u. per ft.-lb. of work = i 2 96 x 778 = °- 00673 

B.t.u. of refrigeration per h.p.-hr. = 0.00673 X 33,000 X 60 = 13,200 

* *- + - u 13200 X 24 • 

Tons of refrigeration per h.p. = OOOO y 144 = 

(Assume 3 lbs. of coal per h.p.-hr.) 
Tons of refrigeration per lb. coal = o v 04 = 0-0153 
Tons of refrigeration per cu. ft. of compressor volume taken in = 
= 0.000235 



144 X 2000 

Gallons of cooling water per cu. ft. of compressor volume taken in for 15 c 
78 6 
rise in cooling water = .. _ 'o or = 0.628 

628 
Gallons per ton of refrigeration = n 0009*3 5 = 2670 

Case II. — Dry compression. 

pi = 38.02 lbs. 

p 2 = 153.90 lbs. 

s 2 = si = 1.1534. 
This is s 2 for 110° F. superheat. 
(This may be checked by 

/r> 2 \ 0.25 /153 9\ 0.25 

r.-T.g) =470( 3 -f!) =664° aba 

Deg. superheat = 664 - 540 = 124° F.) 
i 2 = 628.4 B.t.u. 

M = ^~ = 0.136 

ii = 541.2 
iz = 36.5 

Qc = 0.136 [628.4 - 36.5] = 80.5 
Q r = 0.136 [541.2 - 36.5] = 68.6 
AW = 80.5 - 68.6 = 11.9 
AW with friction = 11.9 X 1.20 = 14.3 
Refrigerating effect without friction yTq = 5.76 

Refrigerating effect with friction r^ = 4.80 

B.t.u. per ft.-lb. of work = u g x 77g = 0.00617 

B.t.u. of refrigeration per h.p.-hr. = 0.00617 X 33,000 X 60 = 12,200 

t t t ■ u 1220 ° X 24 1 no 

Tons of refrigeration per h.p. = 2000 x 144 = 

Tons of refrigeration per lb. ccal (assume 3 lbs. coal per h.p.-hr.) = 
1.02 
3 x2l = 0.0142 



430 HEAT ENGINEERING 

Tons of refrigeration per cu. ft. of compressor volume taken in = 
0.000238 



144 X 2000 

Gallons of cooling water per cu. ft. of compressor volume taken in for 

80 5 
15° rise in cooling water = ,. v o ok = 0.644 

0.644 
Gallons per ton of refrigeration = A qqqo38 = ^700 

Carbon Dioxide. — (Wet compression). 
Tt = T 1 = 470° abs. = 10° F. 
T u = 540° abs. = 80° F. 
T 3 = 525° abs. = 65° F. 
Pi = 362.8 lbs. 
p 2 = 936 lbs. 
i 2 = 80 
s 2 = 0.1511 

Si = 0.1511 = - 0.0226 + xi 0.2164 
xi = 0.80 
i x = - 11.23 + 0.80 X 110.12 

= 76.86 
iz = 21.8 

vi = 0.247 X 0.80 = 0.198 
M = 5.05 

Q c = 5.05[80 - 21.8] = 294 
Q r = 5.05[76.9 - 21.8] = 278.4 
AW = 5.05[80 - 76.9] = 15.6 
AW with friction = 15.6 X 1.50 = 23.4 

(Friction has been increased due to the packing made necessary by the 
excessive pressure.) 

278 4 
Refrigerating effect without friction = -. - '„ = 17.8 

278.4 
Refrigerating effect with friction = ~o \ =11.9 

278 4 
B.t.u. per ft.-lb. work = 2 3 4 x 778 = °' 0131 

B.t.u. of refrigeration per h.p.-hr. = 0.0131 X 33,000 X 60 = 26,000 
rp ( t ■ ,. , 26000 X 24 

Tons of refrigeration per h.p. = 2 Q00 x 144 = 

(Assume 3 lbs. coal per h.p.-hr.) 

2.17 
Tons of refrigeration per lb. coal = „ v 04 = 0-0301 

Tons of refrigeration per cu. ft. of compressor volume taken in = 

278.4 

0.000965 



144 X 2000 

Gallons of cooling water per cu. ft. of compressor volume taken in for 15 c 

294 
rise in cooling water = ,. 5 „_ =2.35 

2 35 
Gallons per ton of refrigeration = n 000Qfi5 ~ ^^^0 



REFRIGERATION 431 

Sulphur Dioxide. — (Wet compression). 

T, = Tx = 470° abs. = 10° F. 

T 2d = 540° abs. = 80° F. 

T 3 = 525° abs. = 65° F. 

pi = 13.5 lbs. 

. p 2 = 59.65 lbs. 

i 2 = 162.1 

s 2 = 0.302 

Si = 0.302 = - 0.1746 + x t 0.3625 

xi = 0.88 

ii = - 6.89 + 0.88 X 169.78 = 142.6 

U = 10.96 

vi = 0.88 X 5.96 = 5.25 

M = 5X5 = ° 19 

Q c = 0.19[162.1 - 10.96] = 28.8 

Q r = 0.19[142.6 - 10.96] = 25.0 

AW = 0.19[162.1 - 142.6] = 3.71 

AW with friction = 3.71 X 1.20 = 4.46 

25 

Refrigerating effect without friction = ^-=r = 6.75 

25 

Refrigerating effect with friction = j-t?; = 5.61 

25 
B.t.u. per ft.-lb. work = 4 46 x 778 = 0.00725 

B.t.u. per h.p.-hr. = 0.00725 X 33,000 X 60 = 14,350 

t e * • *■ v 14350 X 24 

Ions of refrigeration per h.p. = onnn y 144 = 1-195 

(Assume 3 lbs. coal per h.p.-hr.) 

1.195 
Tons of refrigeration per lb. coal = » ' ~ . = 0.0167 

Tons of refrigeration per cu. ft. of compressor volume taken in = 

= 0.000087 



144 X 2000 

Gallons of cooling water per cu. ft. of compressor volume taken in for a 

28 8 
15° rise in cooling water = ., . ' _ = 0.23 
10 X 0.00 

0.23 
Gallons per ton of refrigeration = „ ^Vino^ = 2640 



ABSORPTION APPARATUS 

To eliminate as far as possible the moving parts of a refriger- 
ating apparatus and to utilize waste heat from other machines the 
absorption type of refrigerating machine has been developed. 
In this the High pressure is produced by boiling a solution of NH 3 
in water and driving off* the NH 3 by a steam coil at such a rate that 
the pressure is sufficient to produce a temperature of condensa- 



432 



HEAT ENGINEERING 



tion above the temperature of the supply water. The low 
pressure is maintained by absorbing the vapor in water at a rate 
to produce this pressure. 

The apparatus, Fig. 199, consists of a generator A in which a 
steam coil B supplies enough heat to warm the liquor of aqua 
ammonia to such a temperature that its capacity for ammonia 
is reduced and the ammonia is driven off. This gas is super- 
heated as it leaves the generator and to use some of the heat 
above saturation (as the gas must finally be condensed) it is 
passed over a series of baffle-plates over which the fresh strong 
liquor passes on its way to the generator B. The part of the 
apparatus C is called the analyzer. The function of this is to 

Cooling Water Outlet 
f D E 




Condenser 



Strong 
j£ Liquor 



Fig. 198. — Absorption machine . 



reduce the superheat in the discharge gases. As the gas or vapor 
of ammonia contains some water vapor which would interfere with 
the action of the vapor in the expansion valve, and as it would use 
some of the refrigerating capacity when it absorbs ammonia 
after its condensation, a rectifier D is used to reduce the water 
content. This rectifier cools the vapors still further. The heat 
is absorbed by cold water or by the strong liquor which is pumped 
from the absorber through tubes in the rectifier on its way to 
the generator. This substance is at a lower temperature than 
the vapors, and although low enough to condense most of the 
water it is not low enough to condense the ammonia. The 
condensed water is caught in a separator E and as it absorbs 
ammonia, the aqua liquor thus formed is sent back to the gen- 
erator. From the rectifier the dry vapor is sent into the con- 
denser F where the ammonia vapor is condensed by a cold water 
supply. The liquid is now passed through the expansion valve 



REFRIGERATION 433 

G and enters the expansion coil of the brine cooler H in which 
the liquid is evaporated by the abstraction of heat from the 
surrounding brine. If the direct-expansion system is used this 
heat is drawn from the room. The vapor is drawn into the 
absorber / by the weak liquor which has been allowed to flow 
from the generator. The great capacity of this weak liquor for 
ammonia produces a reduction in pressure of the vapor. This 
absorption of ammonia generates heat and a coil J through which 
cool water is circulated is placed in the absorber to keep the 
temperature at a low point. This low temperature is necessary 
to keep liquor at a point of great capacity for ammonia. The 
absorbing power of water decreases as the temperature is in- 
creased. The strong liquor is pumped from the absorber into 
the generator at higher pressure by the pump K. To save some 
of the heat it is passed through the interchanger L in which 
it takes heat from the hot weak liquor leaving the bottom of 
the generator. The weak liquor flows to the absorber I in which 
the pressure is less than that in the generator. This strong liquor 
is sometimes first passed through the rectifier pipes as it is cool 
enough to condense the water vapor, and from this coil it passes 
to the interchanger or exchanger L and then the strong liquor 
is discharged into the analyzer in which it is still further warmed 
by the superheated vapors. The arrangement in the figure, how- 
ever, is such that water is used to cool the analyzer. The strong 
liquor finally reaches the generator and the boiling drives off 
the ammonia. 

The principle of the absorption machine is the same as that of 
the commpression machine. The vapor is put under such a 
pressure that its temperature of vaporization is above that of a 
water supply and this water may remove heat and condense 
the ammonia; after this it is placed under such a pressure that 
heat may be abstracted from a region of low temperature. The 
pressure is produced by the steam coil in place of the compressor 
and the low pressure is maintained by absorption in place of 
the suction of the compressor. 

PROPERTIES OF AQUA AMMONIA 

The action of ammonia and water (known as aqua ammonia) 
will now be studied to form a basis for the analysis of the action 
of this machine. 

The amount of ammonia or other vapor absorbed by a pound 

28 



434 HEAT ENGINEERING 

of water depends on the pressure and temperature of the water. 
Thus, according to the table prepared by Lucke, water will only 
absorb 3.1 per cent, of its weight of NH 3 at 191° F. and atmos- 
pheric pressure, while at 50 lbs. gauge pressure this temperature 
for the same concentration could be raised to 280° F. 191.5° F. 
is the temperature limit at 50 lbs. pressure for 28 per cent, of 
absorption. 

The relation between the pressure, per cent, of ammonia in 
a solution and the temperature at which this solution will give 
off ammonia or rather boil has been determined experimentally 
by Mollier and although plotted in curves these values are 
represented by Maclntire by the equation 

-t sat. 



Tsol. 



0.00466z + 0.656 (67) 



T sa t. = temperature of saturation of ammonia for a given 

pressure. 
T 80 i. = temperature at which the solution boils. 

x = per cent, of NH 3 in solution = per cent, of con- 
centration. 
= lbs. of NH 3 in 1 lb. of solution. 
This formula holds to 50 per cent, concentration. Above 
that concentration the formula does not give correct results 
but this is not important as such high concentrations are rarely 
used. 

The vapor concentration of the liquid being known with the 
temperature of boiling (T soL ), the temperature (T saL ) cor- 
responding to the boiling pressure, p sa t-, of liquid ammonia is 
found. This gives the total pressure exerted by the ammonia 
and water vapors driven off from the liquor. The partial pressure s 
exerted by the two constituents are not well known at present. 
These partial pressures are proportional to the number of mole- 
cules present or to the partial volumes of each constituent. To 
know how these molecules come off from the solution is a diffi- 
cult problem. Lucke quotes values of partial pressures from 
Perman. These are limited to 140° F. and to concentrations 
from 2.5 to 22.5 per cent, by weight of ammonia. The values 
indicate the manner in which the ammonia vapor pressure 
varies with the water vapor. 

Spangler suggests that the water vapor pressure is equal to 
that for steam at the temperature given multiplied by the ratio 
of number of the molecules of water vapor to the total number 



REFRIGERATION 435 

of molecules present. He assumes that the concentration of 
the vapor is the same as the concentration of the liquid. Thus 
if x is the per cent, solution which is ammonia or the amount 
of ammonia in 100 lbs. of liquor the ratio of the number of 
ammonia molecules and water molecules is 

x ± 100 - x 

Hence if p is the steam pressure for the temperature considered : 

100 - x 
•_ 18 _ 1700 -17 s 

P 100 - x x_ ~ V 1700 + x 
18 + 17 

= partial pressure of moisture. (68) 

This checks with the results of Perman, as given by Lucke, 
with a close degree of approximation. 

If 1 lb. of ammonia be added to a large quantity of water 
it is found that 893 B.t.u. of heat will be liberated. This 
is called the heat of complete absorption. If the ammonia is 
not mixed with a large quantity of water the absorption will 
be partial. If strong ammonia liquor is added to water it is 
found that heat is developed by the solution of this in water; 
the strong solution acting as ammonia. This latter heat is called 
the heat of complete dilution and the difference between the 
heat of complete absorption and that of complete dilution is 
known as the heat of partial absorption. Complete dilution 
occurs when the ammonia is diluted with 200 times its weight 
of water. 

It has been found by Berthelot that the amount of heat de- 
veloped when a liquor of aqua ammonia is diluted so that the 
water content is 200 times the weight of the ammonia, if multi- 
plied by the weight of water per pound of ammonia in the 
original solution gives a constant or practically constant prod- 
uct, 142.5. The heat of complete dilution becomes 

142.5 142.5s 

Q " 100 - x ~ 100 - x (69) 

x 

x — per cent, weight of NH 3 in 1 lb. of mixture. 

Some later results by Thomsen pertaining to much greater 
dilutions than those of the useable experiments of Berthelot's 



436 HEAT ENGINEERING 

indicate that 142.5 may be used, the constant varying from 154 
for 19.27 parts of water, to 148 for 29.36 parts and 113 for 56.33 
parts. 

If now 893 B.t.u. is the heat of complete absorption and 

142. 5x 

tt^t is the heat of complete dilution 

100 — x 

142.5a; 
^93 — inr> '_ — = heat of partial absorption. (70) 

The heat of partial absorption is the heat liberated when 1 lb. 
of ammonia in solution is added to enough water to bring it to 
a concentration of x. 

% 100 x 

If Yq^ lbs. of ammonia are absorbed by — r-^r — lbs. of water 

the heat developed will be 

Q = 893j^ - 100(100 _ x) (71) 

If now ammonia were absorbed to change the concentration 
from x to x', the amount of ammonia then present in the water 
contained in the original pound of solution of strength x would be 

100 X 100-a/ UZ; 

This may be seen from the fact that the amount of water in 

100 _ x 
1 lb. of a solution of strength x is — r^ — and the amount of 

ammonia per pound of water in the second condition is 

x' 

100 - x' 

100 — x 
Hence the amount of ammonia in — r^ — lbs. of water is 

100 - x __*>__ _^1 100 - x 
100 X 100 - x' " 100 X 100 - x' 

The heat generated by the addition of this amount of ammonia 
to bring the solution to strength x' is 

x> 100 - x r _ 142.5*' I . . 

100 X loo^L 893 100 - x'l {7i) 

The heat developed by the addition of ammonia to change 
the strength of the solution from x to x' is therefore 

x' 100 - x r 142.5*' -i x r 142.5* 1 

Q = m x loo-^L 893 _ loo-^J " Tool 893 100 - x\ 

(74) 



REFRIGERA TION 437 

The weight of ammonia added is 

r x' 100 - an _ jc_ 
LlOO X 100 - x'\ 100 ~ 

If the heat is divided by the weight the result will be the heat 
per pound or 

" 142 -4loo^ + lafcJ (75) 



893 



This is the amount of heat liberated when 1 lb. of ammonia 
vapor is absorbed by a liquor of concentration x to change the 
concentration to x' . 

The above expression is true for the absorption of ammonia 
but when 1 lb. of ammonia is driven from a solution to change 
the concentration x' to concentration x not only is the heat 
q necessary but in addition to this, it is necessary to evaporate 
the water vapor present and to superheat the vapors. The 
amounts of these quantities will be found in a problem later. 

The density of the liquor of aqua ammonia varies with the 
amount of ammonia absorbed decreasing with the concentration. 
The specific gravity is given for various concentrations x by 

sp- gr. = 1 - ^>[x - ^ + -^] (76) 

The strong solution would be lighter and would stay at the top 
were it not for diffusion. 

The specific heat of aqua ammonia will be assumed to be unity. 

When a strong liquor of strength x' is changed to strength x 
by giving off 1 lb. of ammonia the number of pounds of strong 
solution, y, required is given by 

1_JL 

_ 100 100 - x 

V — r i . — i _ (77) 

ioo "Too 

The weight of weak solution is (y — 1). 

The above discussion gives the necessary properties of aqua 
ammonia and to apply them as well as to compute the heat trans- 
fers in the various parts of the apparatus a problem will be 
solved. 



438 HEAT ENGINEERING 



PROBLEM 



Find the amount of refrigeration per pound of ammonia driven 
off per minute from 25 per cent, solution in the generator if the 
room temperature is to be held at 25° F. and the cooling water 
varies from 50° to 65° F. How much steam at 20 lbs. gauge 
pressure and x — 0.9 would be required per minute per ton of 
capacity? Find the heat transfer of the various parts of the 
apparatus together with the amount of water used and a heat 
balance. 

Temperatures and Pressures. 

Temperature of ammonia in condenser with a 15° difference 
from the hottest water is 65° + 15° = 80° F. 

Temperature in expansion coil with 15° difference of tempera- 
ture is 25° - 15° = 10° F. 

Temperature of steam in generator (at 34.7 lbs. abs.) = 
259° F. 
Pressure in condenser 

Ammonia pressure (for 80° F.) = 153 . 9 lbs. 
Steam pressure (for 80° F.) = 0.5 1b. 
Total pressure =154. 4 lbs. 

Pressure in expansion coil (10° F.) = 38.02 lbs. 
Pressure in absorber [38.02 — 0.5 (assumed)] = 37.52 lbs. 
Temperature of saturation of ammonia = 9.5° F. 
Pressure in rectifier [154.4 + 0.5] = 154.9 lbs. 
Temperature limit of rectifier (154.9 lbs.) = 80.4° F. 
Pressure in analyzer [154.9 + 0.5] = 155.4 lbs. 
Pressure in generator [155.4 -f- 0.5] = 155.9 lbs. 
Temperature limits in various parts: 

Generator. — For 155.9 lbs. pressure the temperature at which a 
25 per cent, solution will boil is 
80.8 + 460 



T 



0.00466 X 25 + 0.656 



sol- 



T 80l . = j^| = 700° abs. = 240° F. 

This is the lowest temperature possible to drive off the am- 
monia with 25 per cent, concentration. As the evaporation is 
carried on the concentration becomes less. To boil the liquor 
to a smaller concentration requires a temperature higher than 
240° F. The heat required to liberate the ammonia from the 



REFRIGERATION . 439 

strong liquor reduces the temperature of the weak liquor and 
makes the temperature of the vapors leaving the generator that 
required by the formula for the strong liquor. This will be the 
assumption used in this work. 

The limiting concentration at this point when the liquid 
has time to be brought to within 5° of the temperature of steam, 
or 259 - 5 = 254° F., is 

orrf 4 °'Ln = 0.00466z + 0.656 
254 + 460 

102 
x = 77777 = 21.9 per cent. 
4.oo 

This gives a very small change in concentration. Suppose that 
the gauge pressure is raised to 30 lbs. This gives a temperature 
of 274° F. 

274 5 +460 = °- 00466x + °' 656 

x = 17.50 per cent. 
This will be used. 

Thus the pressure in the condenser fixes the pressure in the 
generator and this with the concentration fixes the temperature 
or pressure for the steam used to boil solutions. 

The temperature limit of the absorber is given by 

9.5 + 460 



Tsol. 

469.5 



= 0.00466 X 25 + 0.656 
= 607° abs. = 147° F. 



SOL. Q 773 

Suppose this is kept at 145° F. 

For a 15° difference in temperature in the coils of the absorber 
the water should be kept at 130° F. Of course the water from 
the condenser could be used here as its temperature is not above 
65° F. 

Limiting conditions leaving rectifier are given by the tempera- 
ture of condensation of the ammonia. This is 80.4° F. so the 
temperature at this point may be taken at 90° F. The possible 
concentration at 90° F. and pressure 154.9 lbs. is 

*i£+S! = 0.00466* + 0.656 

x = 70 per cent. 
Having these limiting concentrations, the amount of moisture 



440 . HEAT ENGINEERING 

at the various points may be found except at the discharge from 
the analyzer because at that point the temperature is not known 
as it is due to the mixture of the strong liquor and the liquor 
from the rectifier. To find this, the amount of strong liquor to 
give 1 lb. of ammonia when changed from 25 per cent, concentra- 
tion to 17.5 per cent, concentration is computed. The formula 

_ 100 - x 

y ~ x' - x 

is true for absorption and gives 11.00 lbs. for y for the conditions 
above. 

100- 17.5 11ft _ 

y = ^5^5 = n - 00 

When ammonia is driven off some steam is driven with the 
ammonia and for this reason the concentration of the remaining 
liquor is made greater. To find the value of y the conditions 
leaving the generator must be known. 

Total pressure in generator 155 . 9 lbs. 

Temperature in generator 240° F. 

Concentration 25 per cent. 

(1700 _ 17 y ok\ 
1700 4- 25 — ) = 18 ' 51bs - 

Partial pressure of ammonia 137 . 4 lbs. 

Saturation temperature ammonia 73 . 5° F. 

Superheat 166. 5°F. 

Heat content ammonia 658 . 1 

Specific volume ammonia 3 . 07 

Saturation temperature steam 224° F. 

Superheat 16° F. 

Heat content steam 1162 

Specific volume steam 22 . 

q Q7 
Weight of steam with 1 lb. NH 3 = |^ = . . . . 0. 1395 

The value of y may now be found 

0.252/ - 0.175 [y - (1 + 0.1395)] = 1 
1 - .1995 in ^ K 
y = 0.075 = 10 - 65 

This is weight of liquor at 25 per cent, concentration entering 
the generator. 

The weight of weak liquor leaving is 

(10.65 - 1.1395) = 9.5105 lbs. 

of 17.5 per cent, concentration. This passes to the absorber. 



REFRIGERATION 441 

The amount of ammonia absorbed by bringing this solution to 
a concentration of 25 per cent, is given by 

[9.5105 (1.00 - 0.175) ~ - 9.5105] = 

10.461 - 9.5105 = 0.9505 

This is the amount of ammonia absorbed but there is still a 
further amount absorbed by the water in the rectifier and an- 
alyzer. That leaving the rectifier includes a further amount, 
although small, which is taken over by the condensed moisture 
formed from the moisture leaving the rectifier and passing into 
the condenser and from it into the expansion coils and absorber. 

The amount of water vapor leaving the rectifier must now be 
found. 

Conditions at discharge of rectifier: 

Total pressure 154 . 9 lbs. 

Temperature 90° F. 

Concentration of liquor leaving 70 per cent. 

T 1700 — 17 X 701 

Partial steam pressure 0.696 X — 17004770 — = 9.20 lbs. 

Partial ammonia pressure 154. 7 lbs. 

Temperature saturation of ammonia 80.3° F. 

Superheat ammonia 9 . 7° F. 

Heat content ammonia 564.4 B.t.u. 

Specific volume ammonia 1 . 99 cu. ft. 

Temperature saturation of steam 53° F. 

Superheat steam 37° F. 

Heat content 1100 B.t.u. 

Specific volume 1640 

From this the amount of moisture leaving per pound of am- 
monia is 

1 99 
1640 = ° 0012 

The amount of ammonia absorbed by this amount of water, 
if condensed, to produce a 25 per cent, concentration is 

0.0012 X |f = 0.0004 
75 

Hence for every pound of ammonia absorbed 0.9996 lb. will 
be absorbed by the weak liquor and 0.0004 lb. will be absorbed 
by the water sent in from condenser. 

Since the actual weak liquor absorbs 0.9505 lb. per pound of 
ammonia, the total weight of ammonia absorbed to make the 
strong liquor is 



442 HEAT ENGINEERING 

§|g- 0.9509 lb. 

and the weight of strong liquor is 

9.5105 + 0.9505 + 0.9509 X 0.0016 = 10.4625 

This liquor at 145° F. is passed through the interchanger which 
is supplied with 9.5105 lbs. of weak liquor at 240° F. The heat 
given up by this liquor in reducing its temperature to 150° F. 
by the countercurrent interchangers is equal to 

Q = 9.5105 [240 - 150] X 1 = 856 B.t.u. 

This assumes that the specific heat of the liquor is 1. 

If a 20 per cent, radiation loss is assumed from the interchanger 
the temperature of the strong liquor leaving this apparatus and 
entering the analyzer is 

0.80X 856 , iitg . 
-T04625- + 145 = 21 °- 5 

This strong liquor is mixed with a small amount of liquor of 
higher concentration but at 90° F. from the rectifier. This will 
decrease the temperature although the heat developed by the 
solution of the strong liquor will increase this latter temperature. 
The net effect is to decrease the temperature about 3^°. The 
exact amount of decrease will be found but to get the quantity 
of liquor from the rectifier for first approximation it will be well 
to assume a 3^° drop rather than no drop at all. 

The gases from the analyzer will be brought to the temperature 
of the liquors entering the top of the analyzer by the cooling effect 
of these as the heat of superheat is much less than the heat re- 
quired to raise the temperature of the liquor to that at the desired 
discharge into the generator from the analyzer. The conditions 
of the vapors leaving the analyzer and entering the rectifier are 
fixed by the temperature and pressure at this point. 

Pressure 155 . 4 lbs. 

Temperature assumed 207° F. 

Concentration R^yf ^ = 0.00466* + 0.6561 x =33.2 

T 1700 17 X 33 2~\ 

Partial pressure steam 13.3 X — 1 700 4- 33 9 = ^ '^ 2 ^ s ' 

Partial pressure ammonia 146.7 lbs. 



REFRIGERATION 443 

Temperature saturation ammonia 77 . 2° F. 

Superheat ammonia 129 . 8° F. 

Heat content ammonia 638 . 9 

Specific volume ammonia 2.71 

Temperature saturation steam 187° F. 

Superheat steam 20° F. 

Heat content steam 1149 . 3 

Specific volume steam 45 . 1 

2.71 
The water vapor per pound of ammonia = j^-y = 0.06. 

Since the water vapor leaving the rectifier per pound of am- 
monia is 0.0012, the amount of water removed from the rectifier 
per pound of ammonia leaving the rectifier is 

0.06 - 0.0012 = 0.0588 

In addition to this the amount of moisture associated with 
the ammonia absorbed is also condensed, call this latter M. 
The concentration of the liquor leaving the rectifier for the analy- 
zer is 70 per cent. Hence 

[ [0.0588 + M] ~ ) 0.06 = M 
0.0588 X ^ = M[l - I] 

M = 0.00957 

Hence the moisture condensed per pound of ammonia entering 
the condenser is 

0.0588 + 0.0096 = 0.0684 

and for 0.9505 lb. of ammonia this is 0.065. 
The ammonia absorbed by this is 

0.065 X ~ = 0.1517 

The amount of liquor passing from rectifier is then 

0.1517 + 0.065 = 0.2167 lb. 

This is at a temperature of 90° F. 

On mixing with 10.4625 lbs. of strong liquor at 210.5° F. the 
temperature of the mixture is given by: 

T _ 10.4625X210.5 + 0.2167X90 _ nc n _ 
1 ~ ~ 10.6792 " = 208 -° 5 

The concentration of this mixture is now found. 



444 HEAT ENGINEERING 

Ammonia from strong liquor = 10.4625 X 0.25 = 2 . 61572 

Ammonia from rectifier liquor = =0.15170 

Total ammonia 2 . 76742 

2.76742 
Concentration = ' fi7Q o = 25.913 per cent. 

The temperature of such a concentration at a pressure of 
155.4 lbs. is given by: 

80.6 + 460 = Q 00466 x 25 Q1 + Q 656 

1 sol- 

T soL = 697° abs. = 237° F. 

Since the temperature of the mixture is below this point the 
solution would remain of strength given were it not for the 
heat of solution when the stronger rectifier liquor is diluted in 
the strong liquor. 

To change 10.4652 lbs. of liquor from strength 25 per cent, 
to 25.91 per cent, requires an amount of ammonia equal to 

10.4625 [0.2591 - 0.25] = 0.0955 

This should also equal the ammonia lost by the rectifier liquor 

0.2167 [0.70 - 0.25913] = 0.0955. 

The mixture of a strong solution in a weaker solution de- 
velops heat. This may be considered as the difference between 
the heat developed when the weak solution is made strong and 
that required to weaken the strong solution. 

Heat = 

/25913 , 25\ „ , ,, rtr /70 , 25913^ 



o. M55 [»- 1 4 2 .(SH| +i) _ 893+14 , 5( | + -| )] 

r70_25] 
130 75 J 



51^-^11 



= 0.0955 X 142.5[2.33 - 0.33] 
= 27.22 

This heat is used to raise the temperature of the mixture and 
part may be used to drive off ammonia if the temperature be- 
comes greater than the temperature of solution for the actual 
concentration at the given pressure. The temperature rise for 
27.22 B.t.u. is 

2722 -2 55°F 
10.6792 ~ *' ■ *' 

This gives 208.05 + 2.55 = 210.55° F. 



REFRIGERATION 445 

This is less than 237° F. and hence there is no evaporation. 
If this were higher it would be well to assume an intermediate 
temperature a little above the solution temperature for the 
pressure and concentration and compute the concentration for 
this temperature. Then the amount of ammonia and water 
vapor driven off would be computed together with the heat 
required to do this. If the assumed temperature were correct, 
this heat together with the heat to raise the solution from 208.7 
to the assumed temperature would equal 27.22 B.t.u. If the 
sum is greater, a lower temperature is assumed and the quanti- 
ties computed. After several assumed temperatures the correct 
one may be interpolated. 

In any case the final answer gives the temperature at the dis- 
charge from the analyzer and the second value of condensation 
in the rectifier may be found. The actual conditions at discharge 
from the analyzer into the rectifier are as follows: 

Pressure 155 . 4 lbs. 

Temperature 210.25° F. 

Concentration 32 . 3 

Partial steam pressure 9 . 39 lbs. 

Partial ammonia pressure 146.01 lbs. 

Temperature saturation ammonia 76.9° F. 

Superheat ammonia 133 . 3° F. 

Heat content ammonia 640 . 9 

Specific volume ammonia 2 . 73 

Temperature saturation steam 190° F. 

Superheat steam 20.25° F. 

Heat content steam 1151 

Specific volume steam 42 . 4 

Water vapor per lb. of ammonia 0.0644 

Water condensed per lb. of ammonia passing into condenser 

0.0644 - 0.0012 = 0.0632 

70 
M = 0.0644 [0.0632 + M] ^ 

= 0.01117 

Total moisture = 0.0632 + 0.01117 = 0.0744 per lb. 

Moisture for 0.9509 lb. = 0.0707 

70 
Ammonia absorbed = 0.0707 X^ = 0.165 

Liquor from rectifier = 0.165 + 0.0707 = 0.2357 

t, , , . . 10.4625 X 210.5 + 0.2357 X 90 

Temperature of mixture = 1 o 6QS2 " = 20 ?.84 

2 7807 
Concentration of mixture = 1fl finco = 25.99 per cent. 



446 HEAT ENGINEERING 

Temperature solution = 696° abs. = 236° F. 
Ammonia taken up by weak solution 

10.4625 [0.2599 - 0.25] = 0.1036 

Ammonia given up by strong solution 

0.2357 [0.70 - 0.2599] = 0.1037 

T70 251 
Heat of change of solution 0.1037 39-75 142.5 = 29.55 

29.55 
Temperature rise = i r> aqq o = 2.76° F. 

Temperature of discharge = 207.84 + 2.76 = 210.6° F. 

This is below 236° F., hence there is no further change of con- 
centration and the condition of the entrance into rectifier of 
210.25 may be used as the difference in temperature does not 
change data. 

The ammonia entering rectifier is therefore 

0.9509 + 0.165 = 1.1159 

The vapor entering is given by two methods: 

Vapor = [0.0744 + 0.0012] 0.9509 = 0.0719 
Vapor = 1.1159 X 0.0644 = 0.0740 
Mean value = 0.073 

The liquor at temperature 210.25° F. drops through the 
analyzer and receives heat from the hot vapors. As this heat 
is not sufficient to heat the liquor to 240° F., it will be assumed 
that a steam coil is used to supply the necessary heat which 
will be computed. The liquor is assumed to leave at 240° F. 
The conditions for this are given for exit from generator on 
p. 440. 

Since it is known that the discharge from the generator is 1 lb. 
of ammonia and 0.1395 lb. of steam, at each point of the appa- 
ratus the amount of substance is known. 

This can be put in tabular form : 

Generator: 

Entering. — 10.65 lbs. liquor of 25 per cent, concentration at 240° F. 
Leaving. — 1 lb. ammonia vapor at 240° F. 
0.139 lb. water vapor at 240° F. 
9.501 lbs. of liquor of 17.5 per cent, concentration at 240° F. 

Analyzer: 

Entering. — 1 lb. ammonia vapor at 240° F. 

0.139 lb. of water vapor at 240° F. 

10.67 lbs. liquor of 25.99 per cent, concentration at 210.6° F. 



REFRIGERATION 447 

Leaving. — 10.65 lbs. liquor of 25 per cent, concentration at 240° F. 
1.116 lbs. of ammonia vapor at 210.6° F. 
0.074 lb. of water vapor at 210.6° F. 

Rectifier : 

Entering.— 1.116 lbs. of ammonia at 210.25° F. 

0.074 lb. of water vapor at 210.25° F. 
Leaving. — 0.9509 lb. ammonia vapor at 90° F. 

0.0011 lb. water vapor at 90° F. 

0.236 lb. liquor of 70 per cent, concentration at 90° F. 

Condenser: 

Entering. — 0.9509 lb. ammonia vapor at 90° F. 

0.0011 lb. of water vapor at 90° F. 
Leaving. — 0.9505 lb. of ammonia liquid at 80° F. 

0.0015 lb. liquor of strength 25 per cent, at 80° F. 

Expansion Coil: 

Entering. — 0.9505 lb. ammonia liquid at 80° F. 

0.0015 lb. liquor of strength 25 per cent, at 80° F. 
Leaving. — 0.9505 lb. ammonia at 10° F. 

0.0015 lb. water vapor (strength 25 per cent, assumed) at 10° F. 

Absorber: 

Entering. — 9.501 lbs. liquor of 17.5 per cent, concentration at 150° F. 

0.9505 lb. ammonia at 10° F. 

0.0015 lb. water vapor at 10° F. 
Leaving. — 10.463 lbs. liquor of 25 per cent, concentration at 145° F. 

The heat interchange for each part is found by the difference 
between the intrinsic energy and the work (or heat content) at 
entrance and exit plus an allowance for radiation. 

Generator: 

Entering. — Heat of liquid of 10.65 lbs. 

10.65[240 - 32] = 2218 

Heat of solution of liquor 

-0.25 X 1065 T893- 142.5 X ||1 = - 2255 

-~ 37 
Leaving. — Heat of 1 lb. ammonia = 1 X 658.1 = 658.1 

Heat of 0.1395 lb. steam = 0.1395 X 1162 = 162.0 

Heat of liquid of 9.501 lbs. liquor 

9.501 [240 - 32] = 1975.0 

Heat of solution of liquor 

-0.175 X 9.501 [893 - 142.5 X ^|] = - 1435.0 

1360.1 
Net heat to be supplied = 1360.1 - (- 37)= 1397.1 



448 HEAT ENGINEERING 

Analyzer: 

Entering. — Heat of 1 lb. ammonia = 1 X 658.1 = 658.1 

Heat of 0.1395 lb. steam = 0.1395 X 1162 = 162.0 

Heat of liquid of liquor = 10.67 [210.25 - 32] = 1910.0 

Heat of solution = - 0.2599 X 10.7 ^893 - 142.5 X |^1 = -2340.0 

390.1 
Leaving.— Heat of liquid of 10.65 lbs. = 10.65 [240 - 32] = 2218 

Heat of solution = - 0.25 X 10.65 1893 - 142.5 X |gl = - 2255 

Heat of 1.116 lbs. ammonia = 1.116 X 640.9 = 716.0 

Heat of 0.074 lb. steam = 0.074 X 1151 = 85.2 

764.2 
Net amount added = 764.2 - 390.1 = 374.1 

Total amount added in generator and analyzer allowing 20 
per cent, for radiation is 

120 [1397.1 + 374.1] = 2125 B.t.u. 

Pounds of steam required at 30 lbs. gauge and x = 0.9 is given 
by 

Mst = 0.9 X 927.9 = 2 - 55 lbs ' 
Rectifier : 

Entering. — Heat of 1.116 lbs. ammonia = 716.0 

Heat of 0.074 lb. steam = 85.2 

801.2 

Leaving.— Heat of 0.9509 lb. ammonia = 0.9509 X 564.4 = 536.0 

Heat of 0.0011 lb. steam = 0.0011 X 1100 = 1.2 

Heat of liquid of 0.236 lb. liquor = 0.236 [90 - 32] = 13.7 

Heat of solution = - 0.7 X 0.236^893 - 142.5 X ^1 = - 92.5 

458.4 
Net heat abstracted = 801.2 - 458.4 = 342.8 B.t.u. 

Condenser: 

Entering.— Heat of 0.9509 lb. ammonia = 536.0 

Heat of 0.0011 lb. steam = 1.2 

537.2 

Leaving.— Heat of liquid of 0.9059 lb. ammonia = 0.9509 X 53.6 = 51.0 

Heat of liquid of 0.0011 lb. water = 0.0011 X 48.1 = 0.0 

Heat equivalent of ApV = 0.0 

5i7o 

Net heat abstracted = 537.2 -51 = 486.2 B.t.u. 

Expansion Coil: 

Entering. — Heat of ammonia = 51.0 

Heat of water = 0.0 

51T0 



REFRIGERATION 449 

Leaving. — Heat of ammonia vapor = 0.9505 X 541.2 = 514.0 

Heat of water vapor = 0.0015 [10 - 32] = - 0.0 



^893 - 142.5 X ||] = 



Heat of solution = - 0.004 893 - 142.5 X s= = - 3.4 



510.6 



(Work of liquid neglected.) 

Heat abstracted 510.6 - 51 = 459.6 

Tons of refrigeration per pound of vapor per minute: 

459 - 6 O Q * 

W = 2 - 3 tons 

Absorber : 

Entering.— Heat of liquid of liquor = 9.501[150 - 32] = 1120.0 

Heat of solution of liquor = - 0.175 X 9.501 [862.7] = - 1435.0 
Heat of ammonia = 514.0 

Heat of liquor = — 3.4 

195.6 

Leaving.— Heat of liquid of liquor = 10.463 [145 - 32] = 1180 

Heat of solution = - 0.25 X 10.463 [845.5] = -2215 

- 1035 
Net heat abstracted = 195.6 - (- 1035) = 1230.6 

Interchanger : 

Entering.— Heat of liquid of strong liquor = 10.463 [145 - 32] = 1180.0 
Heat of liquid of weak liquor = 9.501[240 - 32] = 1975.0 



Heat equivalent of work = =~ 155.4 — 37.5 X 



144 X 10.463 _ 
62.5X0.913 b5 



3157.6 



4 3 / 25 2 35 3 \ 

[Density = 1 - ^(m - m + wm ) = 0.913] 

Leaving.— Heat of liquid of strong liquor = 10.463 [210.5 - 32] = 1865 
Heat of liquid of weak liquor = 9.501 [150 - 32] = 1120 

2985 
Heat radiated = 3157.6 - 2985 = 172.6 B.t.u. 

2 65 
The work = 2.65 B.t.u = t^-^ X 2 = 0.125 h.p. for 1 lb. of ammonia 

per minute with 50 per cent, efficiency. 



29 



450 HEAT ENGINEERING 

Heat Balance: 



Generator 

Analyzer 

Rectifier 

Condenser. . . . 
Expansion coil. 

Absorber 

Pump 

Exchanger. . . . 



Taken in 


Given out 


1397.1] 




■ 374.1 .. . 


354.0 


354. 0) 






342.8 




486.2 


459.6 






1230.6 


2.65 






172.6 



Total 2587.4 2586.2 

The cooling water entering the condenser at 50° F. is raised to 
65° F. This could then go to the rectifier where its temperature 
could be raised to 75° F. and finally to the absorber where the 
temperature could be increased to 130° F. The weights of water 
would then be: 

Wt. for condenser per 1 lb. ammonia liberated = .. ' = 32.2 

342 8 
Wt. for rectifier per 1 lb. ammonia liberated = * = 34.3 

Wt. for absorber per 1 lb. ammonia liberated = — -- ' = 22.4 

The amount of water actually needed is 34.3 lbs. 

The surface required in these different sections is found by 
methods of Chapter III. 

The other data computed for the other refrigerating machines 
will be computed for comparison. 

Tons of refrigeration per lb. ammonia = onno v 144 = 0.00159 

459 6X8 
Tons of ice melting capacity per lb. coal = 2 __ o'ooo V 144 = ^-005 

(Assuming 8 lbs. of steam per lb.) 

Tons of ice melting capacity per lb. steam = — ^ — = 0.0006 

120 
The brine pump would require 0.125 X -^r = 0.25 lb. of 

steam per minute for 459.6 B.t.u. of refrigeration or 2.55 lbs. of 
steam in generator. Hence the exhaust from the pump can 
be used easily for part of the steam supply. 

Gallons of cooling water with 15° rise per ton of refrigeration = 
34.3 
8.35 X 0.00159 = 258 °* 



REFRIGERATION 



451 











CM 


CM 






CM 








t^ 


l^ 


© 


O 00 


'O 


U5 




ajnssajd /Acyj 


^' 


T* 


00 


00 CM 


co 


b^ 






i— i 




CO 


CO CO 
CO 


i—i 


CO 














iO 








oo 


00 


05 


OS O 


CO 


05 




ajnssajd q3iH 


00 


00 


0? 


CO CO 


05 


IO 






iQ 


IO 


iO 


IO CO 


iO 


»o 


Pn* 








1— 1 


t-I C5 




r-i 






CO 


r^ 


CO 


CM iO 


»o 


O 





(jq-dq 
jad sqi 08) urea^s 


CO 


co 


•o 


t^ o 


CO 


CO 


*o 
co 


o 
o 


O 

o 


c 


rH CO 

88 


C 


O 

o 


o 

Q 


qj jad uoii'BJa 


o 


o 


o 


o 


o 


-Sujaj jo suojl 


d 


d 


6 


o* o 


o 


d 


^TO'Bd'BO 


o 


o 


o 


o o 


o 


o 


Eh 
< 


Stncqaui aot jo 
uo^ jad ja^-B.w. Sui 


oc 

CN 


o 

CM 


b- 

O 
CM 


O CO 
CM CM 


CO 
CM 


oo 

iO 
CM 


w 


-[OOO JO SUOJfBQ 














(=1 


•at 
uwBjp •'jj' -no jad 


IO 

o 

CO 


CO 
O 
co 


00 
CM 


-* IO 


co 






o 


asu cx joj ja'j'BAv 


o 


O 


CO 


CO CO 


CM 






iO 

Ei 
< 


Suqooo jo sno^r) 


o 


o 


o 


O CM 


O 








CO 


to 


to 


«, e 


CO 






« 


japuq^a 


o 


o 


o 


o o 


o 








O^Ul UAlBjp %j -no 
jadA^io'Bd'BO Sui 


XX XX XX 






£ 


-^qaui a'oi jo suoj, 


iO 
00 


o 
05 


>o 

CO 
CM 


oo »o 
co CO 
CM 05 


O0 

























r—t 


00 










13 


(jq--dq jad -q T g) 


CO 

CO 


CO 

CO 


CO 
iO 


CM i-i 
tH o 


CO 


o 

iO 


o 


[\joo -qj jad -jq f% 


o 


o 




i-i CO 




o 


o 


jad ^■jto'Bd'Bo Sui 


o 


o 


c 


o o 


c 


o 


w 

Eh 


-^iatu aoi jo suoj, 


o 


O 


o 


d o 


o 


d 






Tfl 






iO 






s 


•d-q jad -jq fz 


CO 


CO 


o 


CM 1> 


OS 






jad A^ioBd'Bo 3ui 


<M 


CM 


T— 1 


O i—i 


*"* 






3 
o 
o 


-■jjaui a'oi jo euo j. 


o 


O 


1—1 


i-H CM 


1—1 






Pn 




o 


O 


o 


o o 


o 








*i q 


CM 


00 


o 


o o 


K> 








i— i 




CM 


CM O 


CO 






hh' 


-dq jad uoi^Bja 


co~ co~ co" cm~ co~ ^ 









-Sujaj jo 'h'^'g 






1—1 


r-i CM 


rH 






IO 


















CM 




Cr. 


cm 


CO 


l> 


iO 






O 




IO 


CO 


l> 


i—i i— I 


Ol 






^JOM. JO 

•qj-^j jad uot'jBja 


c 
c 


C 

o 


CO 

8 


CO CO 

O H 

O C 


8 








-Sujaj jo -rivj-g 
















8 




c 


c 


c 


d o 


c 






UOI^DIJJ 


-* 


CO 


CM 


o 










q^iAV abtreuuoj 


<N 


CM 


CM 


00 05 


cC 






-jad jo q.uaioqjaoQ 






iO 


Tt< i-H 


iO 

















r-i 






_ 
















fc 




















£ 




















<J 


9 


















« 


a 


















H 


A 

















© 


a 














.2 




ej 


















s 

a 

o9 
u 

i> 
bD 






CL 




■j 




o 

c3 




< 


u 


'2 


o3 ^ 
O P 


4- 

0. 


" 2 
fl 






a. 


'3 

a 


c 


,£ 


o 






"c 


■ E 
< 




c 


6 


< 





452 HEAT ENGINEERING 

From the above it is seen that C0 2 gives a much larger capacity 
per cubic foot and is much better for efficiency as shown by tons 
of refrigeration per pound of coal or steam. The air machines 
have small capacities per cubic foot and are very inefficient. The 
absorption machine for the conditions shown is not as efficient 
as the compression apparatus although if waste steam may be 
used this has a great value. The high pressure necessary on 
the CO2 machine is the objectionable feature of this. 

TOPICS 

Topic 1. — Sketch and describe the action of an air refrigerating machine. 
Give expressions for the heat on each line. Explain what is meant by per- 
formance or refrigerative effect. 

Topic 2. — -Sketch and describe the action of a refrigerating machine using a 
volatile fluid and its vapor. Sketch the T-S diagram and from it give the 
expressions for the heat on each line. Explain what is meant by perform- 
ance. Explain what is meant by a ton of refrigeration. 

Topic 3. — Sketch and explain action of the absorption refrigerating ma- 
chine. Give the function of each part of this apparatus. 

Topic 4. — Explain method of finding the temperatures and pressures of the 
various points on the air cycle and ammonia cycle of a refrigerating machine. 
What is a dense air machine? Give the expressions for work, heat removed 
by the cooling water in each case and heat removed from the refrigerator. 

Topic 5. — What is the effect of clearance? Show this completely. What 
is the effect of friction? Give the expressions for the power necessary for air 
and ammonia machines per pound of substance used. How is the volume of 
the compressor found? That of the expander? 

Topic 6. — What is the effect of moisture in an air machine? Derive an 
expression for the work in the expander when moisture is present and when 
no moisture is present. Derive expressions for the work of the compressor 
with and without moisture. From the expressions above write the expres- 
sions for work with and without friction. 

Topic 7. — What vapors are used for refrigerating machines ? What are the 
advantages of one over the other? What is meant by wet and dry compres- 
sion? Which is the better? Why? What is the clearance factor? Sketch 
T-S diagrams for each and explain action. Why is the expansion valve used ? 
Write the expressions for work, heat and refrigerating effect. 

Topic 8. — What is aqua ammonia? What is the effect of adding ammonia 
to water? What is the effect of adding aqua ammonia to Water? What fixes 
the pressure at which ammonia will be given off from aqua ammonia at a 
certain temperature? What fixes the partial pressures of the ammonia and 
water vapors discharging from this solution? What is meant by per cent, 
ammonia or per cent, concentration? 

Topic 9. — What is the heat of complete dilution? Complete absorption? 
Partial absorption? What is manner of variation of the density of aqua 
ammonia? What value is assumed for the specific heat of aqua ammonia? 



REFRIGERATION 453 

Topic 10. — Outline the analysis of the action of the absorption machine. 
What data is assumed and what does this fix? 



PROBLEMS 

Problem 1. — An air machine is closed and operates between 40 lbs. gauge 
and 200 lbs. gauge with cooling water at 60° F. to 75° F. and a room held at 
5° F. Find the temperatures at the various corners. Find the horse-power 
per ton of refrigeration with 10 per cent, friction effect. Find the cubic feet 
of displacement per minute in each cylinder and the amount of cooling water 
per minute per ton of refrigeration. Clearance 5 per cent, on each cylinder. 

Problem 2. — An air machine is to operate under conditions of Problem 1 
except that it is open and operates to 45 lbs. gauge pressure. Find the quan- 
tities asked for in Problem 1. 

Problem 3. — An ammonia machine operates with cooling water from 60° F. 
to 75° F. and cools a room with direct expansion to 5° F. Find the pressures 
used. Find the horse-power per ton of refrigeration with 10 per cent, fric- 
tion effect and clearance 3 per cent. Find the amount of water per minute 
and the displacement per minute for 1 ton. (a) Solve this with wet com- 
pression, (b) Solve this with dry compression. 

Problem 4. — Solve Problem 3 using C0 2 as the medium. 

Problem 5. — Solve Problem 3 using SO2 as the medium. 

Problem 6. — A 25 per cent, solution is made by the addition of ammonia 
to a 10 per cent, solution of aqua ammonia. How much ammonia has been 
added per pound of original solution? What is the density of each solution? 
What heat is developed by this addition of ammonia? 

Problem 7. — A 25 per cent, solution is to be boiled at 170 lbs. gauge 
pressure. At what temperature will it boil? At what temperature must 
this be heated if it is to be reduced to a 10 per cent, solution. What are the 
partial pressures above the 10 per cent, solution in this case? 

Problem 8. — Find the amount of heat per pound of vapor in the vapors 
coming from a 10 per cent, solution of aqua ammonia boiling at 170 lbs. 
gauge pressure. 



INDEX 

Authors, Authorities, Inventors and Investigators 



Adams, 210 

American Rad. Co., 101 
Andrew, 360 
Atkinson, 363 
Avogadro, 27 

Ball, 233 

Barnett, 359 

Barr, 209 

Barrus, 178 

Barsanti, 359 

Beau de Rochas, 360 

Bernouilli, 279 

Berthelot, 435 

Bertrand, 39 

Bisschop, 360 

Boltzmann, 73 

Boulvin, 189 

Box, 348 

Boyle, 26 

Brille, 80 

Brown, 359 

Buckingham, 314 

Buffalo Forge Co., 101 

Callendar, 198, 204, 208 

Carcanagues, 80 

Carnot, 9, 11, 12, 13, 62, 170 

Carrier, 343 

Charles, 26 

Clark, 233 

Clausius, 167, 170 

Clayton, 210, 214 

Clerk, 360 

Coker, 369 

Corliss, 167 

Cotterill, 198 

Curtis, 302 

Dalby, 77, 80 
Dalton, 28 



Davis, 174, 317 
De Laval, 301 
Dent, 156 
Diesel, 364 
Duchesne, 209 
Dulong, 73, 403 
Dupre Hertz, 39 
Duhring, 356 

Fliegner, 60 
Foran, 335 

Giffard, 275 

Gilles, 360 

Goodenough, 16, 17, 25, 39, 44, 45, 

47, 215 

Hagemann, 99 
Hall, 205 
Halliday, 80 
Harn, 183 

Hausbrand, 99, 100, 356 
Hautefeuille, 359 
Heck, 199, 223 
Heinrich, 209 
Hilliwell, 360 
Huygens, 359 

Ingersoll Rand Co., 120, 151 

Jelinek, 99 
Jordan, 80, 92 
Josse, 261 
Joule, 2, 37, 100 

Kelvin, 8, 36, 37 
Klebe, 44 
Kneass, 277, 279 
Knoblanch, 44, 45 
Kreisinger, 94, 108 



455 



456 



INDEX 



Langen, 44, 360 
Le Chatelier, 44 
Lenoir, 359 
Linde, 44 
Lucke, 73, 464 

Mallard, 44 

Marks, 174, 198, 317 

Matteucci, 359 

Maxwell, 24, 75 

Mclntire, 434 

McMullen, 96 

Mollier, 45, 47, 80, 99, 100, 434 

Moyer, 274, 292, 303 

Napier, 60 

Newcomen, 167 

Newton, 73, 359 

Nicolson, 80, 89, 90, 91, 198, 204, 208 

Nusselt, 75, 94 

Orrok, 97, 334, 335 
Otto, 360 



Ray, 94, 108 
Reeve, 189 
Reynolds, 77, 78, 167 
Rice, 198 
Richards, 156 
Ruggles, 348 

Sangster, 119 

Savery, 167 

Scoble, 369 

Ser, 80 

Spangler, 177, 344, 396, 434 

Stanton, 80, 91 

Stefan, 73 

Stodola, 269, 270, 271 

Stone, 215 

Street, 359 

Stumpf, 167, 226, 231 

Sultzer, 167 

Thomsen, 435 
Thurston, 198 
Tyndall, 73 



Papin, 167, 359 
Parsons, 305, 
Perm an, 434 
Perry, 78, 198 
Petit, 73 
Porter, 167 

Rankine, 60, 167, 170, 172, 215 
Rateau, 272, 304, 349 



Watt, 167 
Webb, 232 
Werner, 80 
Weymouth, 370 
Willans, 167, 226 
Worcester, 167 
Wright, 359 

Zeuner, 215 



INDEX 

Subjects 



Absorber, 433 
Accumulators,. 348 
Adiabatic, 2, 5, 16, 31, 54 
Adiabatic action, 2 
Afterburning, 368 
Aftercooler, 120 
Air, excess, 404 

motor, 151 

pump, dry, 334 

transmission, efficiency, 152 
loss, 152 
Allowance for turbines, 328 
Ammonia, 428 

aqua, 434 
Analysis, Hirn's, 183 

proximate, 402 

T-S, 183 

ultimate, 402 
Analyzer, 432 

Aqua ammonia, concentration, 434 
density, 437 
heat of absorption, 435 
partial pressure, 434 
properties, 433 
specific heat, 437 
Ash and moisture, free basis, 402 
Availability, 1, 14, 62 

conditions for, 14 

loss of, 115 
Avogadro's law, 27 
Axial thrust, 300 

Back pressure, effect of, 177 
Binary engine, 261 
Blades, number of, 327 
Bleeding engines, 255 
Blowing engines, 119 

tubs, 119 
Bone-Schnable boiler, 407 
Boyle's law, 26 
British thermal unit, 1 



Calorimeters, 178 
Capacities, thermal, 21 
Carbon dioxide, 430 
Cards, combined, 243 

computation for, 243 

construction, 243 
Characteristic equations, 19 
Charles' law, 26 
Charts, I-S, 47 

T-S, 47 
Coils for heating, 101 
Clearance, effect, 124, 417 

effect eliminated, 418 

factor, 125 

ratio, 125 

space, 125 

steam, 170, 184 
Closed system, 412 
Combining tube, 281 
Combustion, 375, 402, 403 

heat, 376 

surface, 406 

table, 377 
Complex cycle, 38 
Compound engine, 241 
Compounding for turbines, pres- 
sures, 297 
velocity, 297 
Compression, best point, 230 

curve, 127 

dry, 425 

method, 128 

temperature, 127 

wet, 425 
Compressors, 118 

air, 118 
Condensation, initial, 196 

maximum, 207 
Condenser, ammonia, 98 

barometric, 337 

contraflo, 337 
457 



458 



INDEX 



Condenser design, 338 

dry tube, 337 

jet, 333 

steam, 97 

surface, 333 

uniflux, 337 
Conduction, 15, 72, 74 

coefficients, 95 

factors, 74 
Conductivity, 207 
Conservation of energy, 2 
Constant steam weight curve, 217 

universal gas, 27 
Consumption, computation, 224 

curves, 222 

steam, 173, 210 
Convection, 72, 75 
Converging nozzle, 273 
Cooling, loss from, 140 

towers, 341 
cost, 347 
size, 347 
Counter flow, 82 

Criterion for exact differentials, 4 
Critical pressure, 267 

temperature, 49 
Cross products, 37 
Curve, construction, 218 

expansion, 215 

pressure in nozzle, 269 

steam consumption, 222 
Curtis turbine, 302, 311 

Dalton's law, 28 

De Laval turbine, 301, 309 

Delivery tube, 282 

shape, 282 
Determination of K, 86 
Diagrams, combined, 240 

factor, 221 

IS, 47 

Mollier, 47, 52 

temperature-entropy, 172, 175 
Differential effect of heat, 43 

exact, 3 
Diffusivity, 207 
Dilution, 368 
Discharge, 59 
Displacement of air compressor, 135 



Double bottoms, 356 
Drip pot, object of, 182 
Dry compression, 425, 429 

basis, 402 

test, 187 
Dulong's formula, 408 
Duty, 67 

Efficiency, 12 

actual, 63 

air standard, 362 

Carnot, 12, 14, 62 

change, 389 

conditions for maximum, 66 

cycle, 175 

equality, 12 

kinetic, 293 • 

maximums, 12, 13, 62, 65, 293 

mechanical, 63 

overall, 63 

practical, 63 

Rankine, 171, 173 

theoretical, 62 

transmission of heat, 103 

type, 63 

volumetric, 126 
Effects, single, double, quadruple, 

350 
End thrust, 306 
Energy, conservation, 2 

high grade, 1 

internal, 2 

intrinsic, 41 
Engine, air, displacement, 146 
power, 146 
work on, 138 

and turbine, combined, 329 
compared, 309 

binary, 261 

bleeding, 255 

heat, 62 

internal combustion, 359 

Mietz and Weiss, 366 

multiple expansion, 240 

Otto, 365 

receiver, 242 

regenerative, 256 

similar to turbine, 309 

size of, 229 



INDEX 



459 



Engine steam, 167 

steam cycle, 170 

straight flow, 170 

Stumpf, 226 

tests, 67, 68, 178 

uniflow, 226 

Woolf, 242 

work, 246 
Entropy, 16, 37, 43, 47, 51 

around cycle, 18 

diagram, 19 
Equality of temperature scales, 36 
Equation, Bernouilli, 279 

characteristic, 19 

differential, 29 

fundamental, 21 
Equivalent evaporation, 410 
Evaporation, equivalent, 410 

factor, 410 
Evaporators, 98, 349, 350, 353 
Exact differential, 3 
Expansion, complete, 138 

curves, construction of, 218 

free, 172, 175 

incomplete, 138, 172, 175, 418 

line, 140, 215 
loss, 140 

ratio, 220 
Exploring tube, 269 
Explosion temperature, 389 
Exponent value, 214 
External work, 2, 31, 41 

Factor diagram, 221 

of evaporation, 410 

of safety, 100 
Fan blowers, 118 

power, 149 
Feedwater heaters, 98 
First law of thermodynamics, 2 
Flow, counter and parallel, 82 

of fluids, 56 
Free air, 123 
Friction factors, 292 
Fuels, blast furnace gas, 374 

coal, 402 

crude oil, 374 

gasoline, 374 

illuminating gas, 370 



Fuels, kerosene, 375 
lignite, 402 
natural gas, 370 
peat, 402 
produce gas, 371 
wood, 402 

Gas composition, 376 

engine, temperature, entropy- 
diagram, 392 
test, 69, 70, 396 

producer, 371-373 
Generator, 432 
Governing, air compressors, 149 

gas engines, 367 

steam engines, 235 

multiple expansion engines, 254 
Graphical representation of heat, 6 

Heat, balance, 402, 450 

content, 37, 42, 47, 51 

engine, 62 

graphical representation, 6 

internal, 41 

of liquid, 40 

of vaporization, 40 

on line, 6, 7, 35 

on path, 20, 42, 52 

transmission, 72, 75, 99, 103, 104 
efficiency, 103 
Hirn's analysis, 183, 260 
High grade energy, 1 
Humphrey compressor, 122 
Hyperbola, rectangular, 55, 217 
Hydraulic radius, 80 

compressor, 119, 122 

Ignition, 368 
Impact coefficient, 284 
Indirect heaters, 101 
Initial condensation, 196 
Injector, 275 

capacity, 284 

density of discharge, 283 

details, 281 

double jet, 284 

experiments, 277 

heat equation, 280 

lifting, 284 



460 



INDEX 



Injector number, 282 

overflowing pressure, 285 
temperature, 285 

size, 282 

steam required, 281 

theory, 280 
Interchanger, 433 
Intercooler, 120, 121, 130 

heat removed, 133 

water for, 133 
from, 134 
Internal energy, 2 

heat of vaporization, 41 
Intrinsic energy, 2, 42, 47, 51 

change, 35 
Irreversible cycles, 15, 18 
Isodynamic, 6, 31, 54 

action, 2 
Isothermal, 7, 10, 31, 55 

Jacket, 121, 128, 253 

effect of, 234 

heat removed by, 128 

saving by, 129 

water for, 130 
Jet, force of, 287 

impulse of, 288 

reaction of, 288 

work of, 289 
Joule's equivalent, 2 

Kinetic energy change, 265 

Lagging, 254 
Latent heat, 21 
Law, 14 

Avogadro's, 27 

Boyle's, 26 

Charles', 26 

Dalton's, 28 . 

First Law of Thermodynamics, 
2 

Second Law of Thermody- 
namics, 14 

Stefan-Boltzmann, 73 
Leakage, effect of, 134 

factor, 125 

in pipes, 143 

of valves, 208 



Lines, thermodynamics, 33, 50 

gas-engine card, 392 
Locomobile, 235 

test of, 68 
Logarithmic diagrams, 153, 396 
Loss curves, 274 

of heat 186, 194 
Low-grade energy, 1 

Mats, 341, 347 
Maxwell's equations, 24 
Mean effective pressure, 219 
Missing quantity, 196 
Mixtures, gases, 28 
Molecular weights, 29 
Moisture, allowances, 328 

effect, 418 
Motors, air, 151 
Mouth of nozzle, 268 
Mouthpiece, 272 
Multistaging, 130 

reasons for, 135 

N-value of, 214 
Nozzle, 265, 268, 271 

converging, 273 

spray, 348 

steam, 281 

velocity, 278 

Open system, 412 
Orifices, 271, 272 
Overexpansion, 268 

Parallel flow, 82 

Parson's turbine, 305, 324 

Partitions, heat transmitted, 104 

Paths, 20 

Perfect gases, 19, 26 

Performance, method of reporting, 67 

Polytropic, 32 

Ponds, cooling, 348 

Potentials, thermodynamics, 24 

Power of motor, 126 

Preheater, 152 

Preheating, gain from, 145 

Pressure, allowances, 328 

change, effect of, 176, 177 

critical, 267 



INDEX 



461 



Pressure drop in pipes, 141 

intermediate, 130 

mean effective, 219 

partial, 28, 334 

and volume of steam, 225 
Producer, gas, 371, 373 
Properties of aqua ammonia, 433 
Proximate analysis, 402 
Pumping engine, test, 69, 258 

Quadruple expansion engine, 241 
Quality, 40 

Radiation, 72 

factors, 74 
Radiators, 100 
Rateau, accumulator, 349 

orifice, 272 

turbine, 304, 323 
Ratio of expansion, 220 
Receiver engine, 242 
Rectangular hyperbola, 55, 217 
Rectifier, 432 

Refrigerating machines, absorption, 
431 

air, 411 

dense air, 416 

displacement, 414, 427 

efficiency, 413 

open and closed, 412 

power, 427 

temperature range, 413 

vapor, 424 
Refrigeration, tons of, 414 

power required, 414 
Refrigerative effect, 427 

engines, 256 

performance, 413 
Refrigerator, 9 
Reheaters, 253 
Reheat factor, 312, 323, 324 
Relations, differential, 23 

trigonometric, 291 
Relative humidity, 342 
Reversible action, 10 
Rieman steam shock, 272 
Rotary blowers, 119 

Safety factor, 100 
Saturated steam, 20 
Scale of temperature, 7 



Second law of thermodynamics, 14 

Shock, steam, 272 

Simple cycle, 37 

Solutions, dilute, 99 

Source, 9 

Specific heat, 21, 30 

superheated steam, 30 
Speed, 233 

Stage compressors, 119 
Steam, 20, 39 

consumption, 173, 210 

curves, 222 

turbines, 287 

velocity, 278 

weight curve, 217 

working, 185 
Stefan-Boltzmann law, 73 
Still, Hodges, 353 
Superheat allowances, 328 

effect of, 176, 177, 234 
Superheated vapor, 44 
Sulphur dioxide, 431 

Taylor hydraulic compressor, 122 
Temperature, after compression, 127 

after explosion, 389 

boiling, 356 

critical, 49 

entropy analysis, 188 
diagram, 392 

gas-engine card, 364, 384 

gradient, 76 

intermediate, 133 

Kelvin's absolute scale, 8 

mean difference, 81 

scale, 7 

equality, 36 
Tests, analyses, 183, 188 

engines, 178 

gas engines, 396 

pumping engine, 258 
Thermal unit, British, 1 

capacities, 21 

relations of, 22 
Thermodynamic lines, 33 
Thermodymanics, fundamental, 1 
Thermometers, constant immersion, 

181 
Throat of nozzle, 268 



462 



INDEX 



Throttling action, 58, 123, 425 

loss from, 144 
Thrust, axial, 300 

end, 306 
Tons of refrigeration, 414 
Total heat, 41 
Transmission constants, 105 

heat, 72 

to boiling water, 99 
Trigonometric relations, 291 
Triple expansion, 241 
Tube exploring, 269 
Tubs, blowing, 119 
Turbine, compared with engine, 309 

combined with engine, 329 

computations, 309 

Curtis, 302 

deLaval, 301 

efficiency, 293, 307 

low pressure, 329 

maximum efficiency, 293 

mixed flow, 329 

Parsons, 305 

pressure, 291 

Rateau, 304 

steam, 167, 287 

test, 69 

velocity, 291 
Turbo compressors, 118 

Ultimate analysis, 402 
Underexpansion, 268 
Universal gas constant, 27 



Vacuum allowance, 328 
Valves, air, 136 

leakage, 208 
Vapor, 39 

Variables, independent, 20 
Velocity, actual, 289 

blade, 289 

discharge, 59, 265 

factors, 292 

injector, 278 

relative, 289 

whirl, 290, 292 
Vento heaters, 101 
Viscous liquids, 99 
Volume and pressure of steam, 

225 
Volumes, partial, 28 
Volumetric efficiency, 126 



Walls, effect of, 204, 369 
heat transmission, 104 

Water velocity, 278 

Weights, molecular, 29 

Wet compression, 425, 428 

Woolf engine, 242 

Work, 31 

compression, 123 
multiple expansion, 246 

Working steam, 170, 185 



Y, value of, 266 



